Proof.
If $x \leadsto x'$ is a specialization of points, then there is a canonical map $\mathcal{F}_{x'} \to \mathcal{F}_ x$ compatible with sections over opens and functorial in $\mathcal{F}$. Since every point of $X$ specializes to a point of $Z$ it follows that $\Gamma (X, \mathcal{F}) \to \Gamma (Z, \mathcal{F}|_ Z)$ is injective. The difficult part is to show that it is surjective.
Denote $\mathcal{B}$ be the set of all quasi-compact opens of $X$. Write $\mathcal{F}$ as a filtered colimit $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ where each $\mathcal{F}_ i$ is as in Modules, Equation (17.19.2.1). See Modules, Lemma 17.19.2. Then $\mathcal{F}|_ Z = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i|_ Z$ as restriction to $Z$ is a left adjoint (Categories, Lemma 4.24.5 and Sheaves, Lemma 6.21.8). By Sheaves, Lemma 6.29.1 the functors $\Gamma (X, -)$ and $\Gamma (Z, -)$ commute with filtered colimits. Hence we may assume our sheaf $\mathcal{F}$ is as in Modules, Equation (17.19.2.1).
Suppose that we have an embedding $\mathcal{F} \subset \mathcal{G}$. Then we have
\[ \Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z) \cap \Gamma (X, \mathcal{G}) \]
where the intersection takes place in $\Gamma (Z, \mathcal{G}|_ Z)$. This follows from the first remark of the proof because we can check whether a global section of $\mathcal{G}$ is in $\mathcal{F}$ by looking at the stalks and because every point of $X$ specializes to a point of $Z$.
By Modules, Lemma 17.19.4 there is an injection $\mathcal{F} \to \prod (Z_ i \to X)_*\underline{S_ i}$ where the product is finite, $Z_ i \subset X$ is closed, and $S_ i$ is finite. Thus it suffices to prove surjectivity for the sheaves $(Z_ i \to X)_*\underline{S_ i}$. Observe that
\[ \Gamma (X, (Z_ i \to X)_*\underline{S_ i}) = \Gamma (Z_ i, \underline{S_ i}) \quad \text{and}\quad \Gamma (X, (Z_ i \to X)_*\underline{S_ i}|_ Z) = \Gamma (Z \cap Z_ i, \underline{S_ i}) \]
Moreover, conditions (1) and (2) are inherited by $Z_ i$; this is clear for (2) and follows from Topology, Lemma 5.23.5 for (1). Thus it suffices to prove the lemma in the case of a (finite) constant sheaf. This case is a restatement of Lemma 59.82.3 which finishes the proof.
$\square$
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