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. On the other hand, I must concede that everybody (myself included) calls $\mathcal F'$ a *constant* sheaf. This terminology also seems to me a little misleading but it is firmly entranched now and is here to stay.