Dear Daniel, the reason you couldn't find a proof of your statement nor locate one in the literature is that it is false ; so you were quite right to "have doubts now" ! Here are two formulations of the falseness of the statement.
I) Given a presheaf $\mathcal F$ on a topological space, it is not true that there exists a non-empty open subset $U\subset X$ such that the restriction $\mathcal F |U$ is a sheaf.
For example take $X=\mathbb R$ and $\mathcal F(V)= \mathbb Z$ for all open $V\subset \mathbb R$ (constant presheaf). Since every open $U$ contains disjoint open subsets, the restriction $\mathcal F |U$ is never a sheaf.
II) Given a presheaf $\mathcal F$ on a topological space and its sheafification
$\mathcal F \to \mathcal F'$ it is not true that there exists a non-empty open subset $U\subset X$ such that the restricted morphism $\mathcal F |U \to \mathcal F'|U$ is an isomorphism .
In the preceding example the sheafification $\mathcal F'$ is the sheaf of locally constant $\mathbb Z $-valued functions and again for every $U\subset \mathbb R$ you will find disjoint open intervals $I_1,I_2 \subset U$ for which $\mathcal F(I_1\sqcup I_2)= \mathbb Z \neq \mathcal F'(I_1\sqcup I_2)= \mathbb Z^2$ . So the restricted morphism $\mathcal F |U \to \mathcal F'|U$ is not an isomorphism.
Conclusion I find it ambiguous, as proved by this very question, to call a morphism of sheaves a "local isomorphism" if it is an isomorphism on the stalks. I don't know how widespread this usage is but in my opinion people using it should warn their readers if they decide to adopt it.