**Proof.**
Let $Y = Y_0 \amalg Y_1$ be a decomposition into nonempty open and closed subschemes. We claim that $f(X)$ meets both $Y_ i$. Namely, if not, say $f(X) \subset Y_1$, then we can consider the finite étale morphism $V = Y_1 \to Y$. This is not an isomorphism but $V \times _ Y X \to X$ is an isomorphism, which is a contradiction.

Suppose that $X = X_0 \amalg X_1$ is a decomposition into open and closed subschemes. Consider the finite étale morphism $U = X_1 \to X$. Then $U = X \times _ Y V$ for some finite étale morphism $V \to Y$. The degree of the morphism $V \to Y$ is locally constant, hence we obtain a decomposition $Y = \coprod _{d \geq 0} Y_ d$ into open and closed subschemes such that $V \to Y$ has degree $d$ over $Y_ d$. Since $f^{-1}(Y_ d) = \emptyset $ for $d > 1$ we conclude that $Y_ d = \emptyset $ for $d > 1$ by the above. And we conclude that $f^{-1}(Y_ i) = X_ i$ for $i = 0, 1$.

It follows that $f^{-1}$ induces a bijection between the set of open and closed subsets of $Y$ and the set of open and closed subsets of $X$. Note that $X$ and $Y$ are spectral spaces, see Properties, Lemma 28.2.4. By Topology, Lemma 5.12.10 the lattice of open and closed subsets of a spectral space determines the set of connected components. Hence $\pi _0(X) \to \pi _0(Y)$ is bijective. Since $\pi _0(X)$ and $\pi _0(Y)$ are profinite spaces (Topology, Lemma 5.22.5) we conclude that $\pi _0(X) \to \pi _0(Y)$ is a homeomorphism by Topology, Lemma 5.17.8. This proves (1). Part (2) is immediate.
$\square$

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