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 (essentially equivalent) statements that hopefully clarify the situation.
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 define the presheaf $\mathcal F$ by $\mathcal F(V)= \mathbb Z$ for all open
$V\subset \mathbb R$ (constant presheaf on $\mathbb R$ with values in $\mathbb Z$). 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 of presheaves.
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 of presheaves.
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 a little misleading but it is firmly entranched now and is here to stay.
An answer to Roy's question He asks (in his answer below) for an example of a presheaf all of whose restrictions to open subsets are non-separated. [Recall that a presheaf
$\mathcal F$ is said to be separated if given a covering $U=\cup U_i $ of an open set $U$ by open subsets $U_i$, you can deduce for two sections $f,g\in \mathcal F (U)$ that $f=g$ as soon as you know that $f| U_i=g| U_i$ for all $i$ . This is equivalent to saying that, if $\mathcal F'$ denotes the sheafification of $\mathcal F$, all morphisms
$\mathcal F (U) \to \mathcal F'(U)$ are injective.]
Here is the example. On the topological space $\mathbb R$ consider the sheaf of continuous functions $\mathcal C$, its subpresheaf $\mathcal C_b $ of continuous bounded functions ( Caution: this is not a sheaf !) and the quotient presheaf
$\mathcal F= \mathcal C / \mathcal C_b $ i.e. for $V$ open in $\mathbb R$, $\mathcal F(V)=\mathcal C (V)/ \mathcal C_b (V) $. It is then clear that for all non-empty open $V\subset \mathbb R$ we have $\mathcal F(V) \neq 0$ but for the sheafification $\mathcal F'$ we have $\mathcal F'(V)= 0$ (because every continuous function is locally bounded !). And this is the example requested by Roy: for every non-empty $U$ the restriction $\mathcal F |U $ is a non-separated presheaf on $U$ : $\mathcal F |U \neq 0$ certainly does not inject into $\mathcal F'|U =0$