I am reading Bredon's Sheaf Theory. After defining a Sheaf $\mathcal{A}$ over a space $X$ the author has stated some basic properties. I am stuck with the following statement:
Problem:
For any two sections $s \in \mathcal{A}(U)$ and $t \in \mathcal{A}(V)$, $U$ and $V$ open, then the set $W:= \{x \in U \cap V: s(x) = t(x)\} \subseteq X$ is open in $X$. Further, if $\mathcal{A}$ is Hausdorff then $W$ is closed in $X$.
I have thought about it in the following direction:
Since $U$ and $V$ are open in $X$, $s(U)$ and $t(V)$ are open in $\mathcal{A}$.
Since $\pi$ is a local homeomorphism, we have that $\pi: \mathcal{A} \to X$ is an open map.
$s(W) = t(W)$. If $s(W)$ is open in $\mathcal{A}$, then $\pi(s(W)) = W$ becomes open in $X$. Therefore proving $s(W)$ is open in $\mathcal{A}$ does the job.
Is there some other way to prove the above claim? Or can we prove that $s(W)$ is open in $\mathcal{A}$? Please provide me with some hints, so that I can proceed.