Please allow me to rewrite Will's fantastic answer below, which I hope to be easier to understand and for others' convenience.

Let $(\mathscr{C}, \tau)$ be a Grothendieck site, where $\mathscr{C}$ is a category of schemes and $\tau$ is a topology on $\mathscr{C}$ finer than the Zariski topology, all of whose covers are collections of morphisms that are open for the Zariski topology and surjective on points. Lat $A$ be a set and $\underline{A}$ the associated constant sheaf on $\mathscr{C}$ with the Zariski topology. Then $\underline{A}$ already a (constant) sheaf on $(\mathscr{C}, \tau)$.

Proof. Let $Y\xrightarrow{f}X$ be a $\tau$-cover on the site $\mathscr{C}$. Note that $\underline{A}(X)$ is the set of locally constant functions with values in $A$, hence can be identified with all disjoint $A$-indexed open covers of $X$ (given by mapping $v\in\underline{A}(X)$ to the family $\{v^{-1}(a)\}_{a\in A}$).

Given any $u\in\underline{A}(Y)$ with $up_1=up_2$, we want to find a (unique) locally constant function $v\in\underline{A}(X)$ with $vf=u$. Of course, we have to define $v$ by letting $v^{-1}(a)=f(u^{-1}(a))$ (note that $f$ is an open map), provided that it is well-defined, that is, $f(u^{-1}(a))\cap f(u^{-1}(b))=\varnothing, \forall a\ne b\in A$.

We show this now: Otherwise, there would exist $x\in X$ and $y_1, y_2\in f^{-1}(x)$ with $u(y_1)=a, u(y_2)=b$. By scheme theory, there exists $z\in Y\times_XY$ with $p_1(z)=y_1, p_2(z)=y_2$, thus $a=u(y_1)=up_1(z)=up_2(z)=u(y_2)=b$, a contraction.