Every proof I see that the support of a section $s$ of some sheaf $\mathcal{F}$ is closed contains some line such as "because $s_p \neq 0$ there is some open set $U$ such that $s_q \neq 0$ for all $q \in U$." But, surely a function can have isolated zeros?
Shouldn't the support of $x+3$ on $\mathbb{A}^1$ be the open set $\mathbb{A}^1/\{3\}$? I think I missing something about germs here.
Given a scheme $X$, a sheaf $\mathcal F$ on $X$, a section $f \in \mathcal F(U)$ and a point $x \in X$, there are two notions of $f$ being zero at $x$.
The first notion is the analogue of a function vanishing at a point: $f_x \in \mathcal m_x \subset \mathcal O_{X,x}$, where $\mathcal m_x$ is the maximal ideal. The set of all such $x$ is closed and is often denoted by $X_f$ in the literature and indeed referred to as the "zero locus of $f$". This imitates the fact that the set of zeroes of a polynomial is closed under the Zariski topology.
The second notion (The one in your question) is asking whether the germ of $f$ is zero, i.e. $f_x = 0$. The set of all such $x$ is clearly open by the very definition of a stalk and the complement is called the support of $f$ and is closed.
You should take $X=\mathbb A^1_k = \operatorname{Spec} k[x]$ ($k$ algebracially closed for starters) and $f \in \mathcal O_{\mathbb A^1_k}= k[x]$ to check that the first notion indeed gives you the zeroes of $f$, while the second notion does not give you anything in this case: The support of any non-zero section on $\mathbb A^1_k$ is all of $\mathbb A^1_k$.