Say $A$ is an abelian Banach algebra and $\Omega(A)\subset A^* := B(A, \mathbf{C})$ is the set of non-zero algebra homomorphisms. Consider now $\Omega(A)$ with the weak$^*$ topology.
I am trying to understand why $\Omega(A) \subset S$, where $S$ is the unit ball in $A^*$, is a weak$^ *$ closed subset of $S$ when $A$ is unital. The thing that bugs me is that I think that $\Omega(A)$ is already a closed subset of $S$, without the assumption that $A$ is unital.
The proof that I have in mind is just by considering a net $(\phi_{\lambda})_{ \lambda \in I } \subset \Omega(A)$ that converges in the weak$^ *$ topology to some $\phi \in S$. We want to show then that for any $a, b \in A$ we have that $\phi(ab) - \phi(a)\phi(b) = 0$.
Consider then the following estimate $$|\phi(ab) - \phi(a)\phi(b)| \leq |\phi(ab) - \phi_{\lambda}(ab)| + |\phi_{\lambda}(a)| \cdot |\phi_{\lambda}(b) - \phi(b)| + |\phi(b)| \cdot |\phi_{\lambda}(a) - \phi(a)|. $$ Now we can bound $|\phi_{\lambda}(a)| \leq ||a||$ as $||\phi_{\lambda}|| \leq 1$; similarly $|\phi(b)| \leq ||b||$. It follows by the pointwise net convergence of the $\phi_{\lambda}$'s that we can make the righthand side arbitrary small.
I cannot seem to find a flaw in this reasoning. Let me know if there is something fundamentally wrong. There are related threads to this question see here, or here, but none of them address this particular problem.
I think the comment may have answered your question, but it boils down to we know that each character (when $A$ is unital) $\varphi$ is actually a unital $*$-homomorphism, so in particular, if $\psi_i \longrightarrow \psi \in S$ is a net of characters converging $w^{*}$ to some $\psi \in S$, then by definition of $w^{*}$-convergence: $$\psi(1_{A}) = \lim_i \psi_i(1_{A}) = \lim_i 1 = 1 \neq 0$$ so $\psi \neq 0$ and thus combined with your above argument shows $\psi \in \Omega(A)$.
If you don't assume $A$ is unital, you can get a net of characters tending to $0$ $w^{*}$. For example, Take $A = C_0(\mathbb{R})$. Then $(\text{ev}_n)^{\infty}_{n=1} \subset \Omega(A)$ by $\text{ev}_n(f) = f(n)$. Then $\text{ev}_n \rightarrow 0 \notin \Omega(A)$ $w^{*}$.