**Proof.**
By Sites, Lemma 7.32.7 there is a one to one correspondence between points of the site and points of the associated topos, hence it suffices to prove (1). By Sites, Proposition 7.33.3 the functor $u$ has the following properties: (a) $u(S) = \{ *\} $, (b) $u(U \times _ V W) = u(U) \times _{u(V)} u(W)$, and (c) if $\{ U_ i \to U\} $ is an étale covering, then $\coprod u(U_ i) \to u(U)$ is surjective. In particular, if $U' \subset U$ is an open subscheme, then $u(U') \subset u(U)$. Moreover, by Sites, Lemma 7.32.7 we can write $u(U) = p^{-1}(h_ U^\# )$, in other words $u(U)$ is the stalk of the representable sheaf $h_ U$. If $U = V \amalg W$, then we see that $h_ U = (h_ V \amalg h_ W)^\# $ and we get $u(U) = u(V) \amalg u(W)$ since $p^{-1}$ is exact.

Consider the restriction of $u$ to $S_{Zar}$. By Sites, Examples 7.33.5 and 7.33.6 there exists a unique point $s \in S$ such that for $S' \subset S$ open we have $u(S') = \{ *\} $ if $s \in S'$ and $u(S') = \emptyset $ if $s \not\in S'$. Note that if $\varphi : U \to S$ is an object of $S_{\acute{e}tale}$ then $\varphi (U) \subset S$ is open (see Proposition 59.26.2) and $\{ U \to \varphi (U)\} $ is an étale covering. Hence we conclude that $u(U) = \emptyset \Leftrightarrow s \in \varphi (U)$.

Pick a geometric point $\overline{s} : \overline{s} \to S$ lying over $s$, see Definition 59.29.1 for customary abuse of notation. Suppose that $\varphi : U \to S$ is an object of $S_{\acute{e}tale}$ with $U$ affine. Note that $\varphi $ is separated, and that the fibre $U_ s$ of $\varphi $ over $s$ is an affine scheme over $\mathop{\mathrm{Spec}}(\kappa (s))$ which is the spectrum of a finite product of finite separable extensions $k_ i$ of $\kappa (s)$. Hence we may apply Étale Morphisms, Lemma 41.18.2 to get an étale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that

\[ U \times _ S V = U_1 \amalg \ldots \amalg U_ n \amalg W \]

with $U_ i \to V$ an isomorphism and $W$ having no point lying over $\overline{v}$. Thus we conclude that

\[ u(U) \times u(V) = u(U \times _ S V) = u(U_1) \amalg \ldots \amalg u(U_ n) \amalg u(W) \]

and of course also $u(U_ i) = u(V)$. After shrinking $V$ a bit we can assume that $V$ has exactly one point lying over $s$, and hence $W$ has no point lying over $s$. By the above this then gives $u(W) = \emptyset $. Hence we obtain

\[ u(U) \times u(V) = u(U_1) \amalg \ldots \amalg u(U_ n) = \coprod \nolimits _{i = 1, \ldots , n} u(V) \]

Note that $u(V) \not= \emptyset $ as $s$ is in the image of $V \to S$. In particular, we see that in this situation $u(U)$ is a finite set with $n$ elements.

Consider the limit

\[ \mathop{\mathrm{lim}}\nolimits _{(V, \overline{v})} u(V) \]

over the category of étale neighbourhoods $(V, \overline{v})$ of $\overline{s}$. It is clear that we get the same value when taking the limit over the subcategory of $(V, \overline{v})$ with $V$ affine. By the previous paragraph (applied with the roles of $V$ and $U$ switched) we see that in this case $u(V)$ is always a finite nonempty set. Moreover, the limit is cofiltered, see Lemma 59.29.4. Hence by Categories, Section 4.20 the limit is nonempty. Pick an element $x$ from this limit. This means we obtain a $x_{V, \overline{v}} \in u(V)$ for every étale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that for every morphism of étale neighbourhoods $\varphi : (V', \overline{v}') \to (V, \overline{v})$ we have $u(\varphi )(x_{V', \overline{v}'}) = x_{V, \overline{v}}$.

We will use the choice of $x$ to construct a functorial bijective map

\[ c : |U_{\overline{s}}| \longrightarrow u(U) \]

for $U \in \mathop{\mathrm{Ob}}\nolimits (S_{\acute{e}tale})$ which will conclude the proof. See Lemma 59.29.7 and its proof for a description of $|U_{\overline{s}}|$. First we claim that it suffices to construct the map for $U$ affine. We omit the proof of this claim. Assume $U \to S$ in $S_{\acute{e}tale}$ with $U$ affine, and let $\overline{u} : \overline{s} \to U$ be an element of $|U_{\overline{s}}|$. Choose a $(V, \overline{v})$ such that $U \times _ S V$ decomposes as in the third paragraph of the proof. Then the pair $(\overline{u}, \overline{v})$ gives a geometric point of $U \times _ S V$ lying over $\overline{v}$ and determines one of the components $U_ i$ of $U \times _ S V$. More precisely, there exists a section $\sigma : V \to U \times _ S V$ of the projection $\text{pr}_ U$ such that $(\overline{u}, \overline{v}) = \sigma \circ \overline{v}$. Set $c(\overline{u}) = u(\text{pr}_ U)(u(\sigma )(x_{V, \overline{v}})) \in u(U)$. We have to check this is independent of the choice of $(V, \overline{v})$. By Lemma 59.29.4 the category of étale neighbourhoods is cofiltered. Hence it suffice to show that given a morphism of étale neighbourhood $\varphi : (V', \overline{v}') \to (V, \overline{v})$ and a choice of a section $\sigma ' : V' \to U \times _ S V'$ of the projection such that $(\overline{u}, \overline{v'}) = \sigma ' \circ \overline{v}'$ we have $u(\sigma ')(x_{V', \overline{v}'}) = u(\sigma )(x_{V, \overline{v}})$. Consider the diagram

\[ \xymatrix{ V' \ar[d]^{\sigma '} \ar[r]_\varphi & V \ar[d]^\sigma \\ U \times _ S V' \ar[r]^{1 \times \varphi } & U \times _ S V } \]

Now, it may not be the case that this diagram commutes. The reason is that the schemes $V'$ and $V$ may not be connected, and hence the decompositions used to construct $\sigma '$ and $\sigma $ above may not be unique. But we do know that $\sigma \circ \varphi \circ \overline{v}' = (1 \times \varphi ) \circ \sigma ' \circ \overline{v}'$ by construction. Hence, since $U \times _ S V$ is étale over $S$, there exists an open neighbourhood $V'' \subset V'$ of $\overline{v'}$ such that the diagram does commute when restricted to $V''$, see Morphisms, Lemma 29.35.17. This means we may extend the diagram above to

\[ \xymatrix{ V'' \ar[r] \ar[d]^{\sigma '|_{V''}} & V' \ar[d]^{\sigma '} \ar[r]_\varphi & V \ar[d]^\sigma \\ U \times _ S V'' \ar[r] & U \times _ S V' \ar[r]^{1 \times \varphi } & U \times _ S V } \]

such that the left square and the outer rectangle commute. Since $u$ is a functor this implies that $x_{V'', \overline{v}'}$ maps to the same element in $u(U \times _ S V)$ no matter which route we take through the diagram. On the other hand, it maps to the elements $x_{V', \overline{v}'}$ and $x_{V, \overline{v}}$ in $u(V')$ and $u(V)$. This implies the desired equality $u(\sigma ')(x_{V', \overline{v}'}) = u(\sigma )(x_{V, \overline{v}})$.

In a similar manner one proves that the construction $c : |U_{\overline{s}}| \to u(U)$ is functorial in $U$; details omitted. And finally, by the results of the third paragraph it is clear that the map $c$ is bijective which ends the proof of the lemma.
$\square$

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