I'm stuck with the following fake proof that every compact Hausdorff space $X$ is separable. I don't see where the problem lies.
$X$ is homeomorphic to the spectra $\sigma(C(X))$ of $C(X)$ via the Gelfand transform, where the latter is endowed with the weak* topology. If $A\subseteq C(X)$ is a countable set, the separable $*$-algebra $\bar{A}$ generated by $A$ is such that $\sigma(\bar{A})\supseteq\sigma(C(X))$ (any $*$-homomorphism of $C(X)$ onto $\mathbb{C}$ remains a $*$-homomorphism when restricted to $\bar{A}$). Since $\bar{A}$ is a separable Banach space, the unit ball of its dual is separable and compact in the weak* topology. But $\sigma(\bar{A})$ is a compact subset of this unit ball which is separable and compact, and $\sigma(C(X))$ is a compact subset of $\sigma(\bar{A})$. Therefore $\sigma(C(X))$ is separable and compact, and so is $X$.
I suspect the problem lies in the assertion that $\sigma(\bar{A})\supseteq\sigma(C(X))$, but I cannot see why this is false.
It is not correct that $\sigma(C(X))$ is a subspace of $\sigma(\bar{A})$. Any character on $C(X)$ can be restricted to a character on $\bar{A}$, and this defines a continuous map $\sigma(C(X))\to\sigma(\bar{A})$. However, this map will typically not be injective. (In fact, it follows from Gelfand duality that it is never injective unless $\bar{A}$ is all of $C(X)$.)
It may be instructive to consider the case where $A$ is empty, so $\bar{A}$ is just $\mathbb{C}$. Then $\sigma(\bar{A})$ is a point, and the restriction map $\sigma(C(X))\to\sigma(\bar{A})$ is just the unique constant map.
More generally, $\sigma(\bar{A})$ can naturally be identified with the quotient of $X$ by the equivalence relation which identifies two points $x,y\in X$ if $f(x)=f(y)$ for all $f\in A$.