Suppose that $A$ is a natural Banach function algebra on $K$, a compact Hausdorff space. So $A$ is realised as an algebra of continuous functions on $K$, is a Banach algebra for some norm (necessarily dominating the supremum norm) and each character on $A$ is given by evaluation at a point of $K$.
If $F\subseteq K$ is closed, then how can we prove that $I(F)=\{f\in A\mid f(k)=0,\, k\in F\}$ is also closed?
I thought I'd make the comments by Daniel Fischer into an answer. Here goes:
All that remains to be done now is to show that $I(x)$ is closed. We show $I^c (x)$ is open. To this end, let $f \in I^c(x)$. Then $f(x) \neq 0$. Since the norm on $A$ dominates $\|\cdot\|_\infty$, we know that there exists $C \in \mathbb R$ such that $\|f\|_\infty \le C \|f\|$ for all $f\in A$. Hence if we choose $\delta = {|f(x)| \over 2C}$ and consider $g \in B_{\|\cdot\|}(f,\delta)$ then $$ |g(x)-f(x)| \le \|g - f \|_\infty \le C\|g-f\| < {|f(x)| \over 2}$$ and hence $|g(x)| \ge {|f(x)| \over 2} > 0$ which shows that $g \in I^c (x)$ and hence $B_{\|\cdot\|}(f,\delta) \subseteq I^c (x)$.