How to get the closure of an algebra when the algebra does not vanish nowhere?

Question

If $\mathscr{A}$ is an algebra of continuous real-valued functions on a compact set $K$ that separates points, then either

1. $\bar{A}=\{f: K \rightarrow \mathbb{R} \mid f$ is continuous $\}$ or
2. $\overline{\mathscr{A}}=\{f: K \rightarrow \mathbb{R} \mid f$ is continuous and $f(p)=0\}$ for some $p \in K$.

By Stone-Weierstrass theorem, if $A$ vanishes nowhere, the (1) condition will hold, the closure of $A$ is all the continuous function in $K$. If $A$ does not vanish no where, it means that there exist a $x\in K$, such that $g(x)$ not equal to $0$ for every $g\in A$, does it go directly to condition (2), how to show that in this condition, the closure of $A$ is (2).

thanks!