Let $\mathcal{F}$ be a sheaf of abelian groups over a top. space $X$. Assume also that the sheaf topology is Hausdorff. Then for a section $s: U \to \mathcal{F}(U)$, its support ${\rm supp}(s) = \{x \in U \mid s(x) \neq 0 \}$ is open in $\mathcal{F}$.
My ideas:
The Hausdorff-condition of a sheaf is equivalent to the "Identity Theorem", meaning if for an open set $V \subset U$ $s \restriction_{V} = 0$, then $s = 0$ on all of $U$. But I don't see how this leads to the support being open.
$\def\sF{\mathcal{F}} \def\sO{\mathcal{O}} \def\bbZ{\mathbb{Z}} \def\sh{\mathsf{Sh}} \def\ab{\mathsf{Ab}} \def\supp{\operatorname{supp}} \def\sB{\mathcal{B}} $Let $(X,\sO_X)$ be a ringed space and let $\sF$ be a sheaf of $\sO_X$-modules such that the étale space $|\sF|$ is Hausdorff, and let $s\in\sF(U)$ be a section over an open subset $U\subset X$. We are going to show that $\supp s\subset X$ is open (you can verify that if $\sO_X=\underline{\bbZ}$ is the $\bbZ$-constant sheaf on $X$, then the forgetful functor $\sh_{\underline{\bbZ}}(X)\to\sh_{\ab}(X)$ from sheaves of $\underline{\bbZ}$-modules to sheaves of abelian groups is an isomorphism of categories and, hence, your situation is a particular example of the general case. See e.g. this or this).
For any section $t\in\sF(V)$, denote \begin{align*} \dot{t}:V&\to|\sF|\\ x&\mapsto t_x \end{align*} to the induced continuous section.
Let $x\in\supp s$. Then $s_x\neq 0$. Pick a neighborhood system $\sB$ of $U$ at $x$. Then $\dot{s}(\sB)=\{\dot{s}(B)\mid B\in\sB\}$ and $\dot{0}(\sB)$ are neighborhood systems of $|\sF|$ at $s_x$ and at $0_x$, respectively. Since $|\sF|$ is Hausdorff, there are $B,C\in\sB$ with $\varnothing=\dot{s}(B)\cap\dot{0}(C)=\dot{s}(B\cap C)\cap\dot{0}(B\cap C)$. In particular, $s_y\neq 0_y$ for all $y\in B\cap C$. Thus, $x\in B\cap C\subset\supp s$.
If $s_x \neq 0$ then there is some open set $V$ containing $x$ such that $s _V \neq 0$. By the identity theorem $s$ is nonzero on every open set inside $V$ and hence $s_y \neq 0$ for each $y \in V$ so $V \subset \mathrm{supp}(\mathcal{F})$ proving that the support is open.
