Proof. Let $\fB \subset 2^{X}$ be a base for the topology on $\beta X$, then for each $A \subset \beta X$ clopen, there exists $\mathcal{F}\subset \fB$ finite such that $A = \bigcup_{U \in \mathcal{F}}U$. Should $\fB$ be countable, then there are at most countably many clopen subsets of $\beta X$.

However, for each $A \subset X$, let $U_{A} = \bracsn{\fU \in \beta X|A \in \fU}$, then $\bracsn{U_A|A \subset X}$ represents uncountably many clopen sets. Therefore $\fB$ cannot be countable, $\beta X$ is not second countable.

If $\beta X$ was metrisable, then it is second countable by compactness and Proposition 8.1.2. Therefore $\beta X$ is not metrisable.$\square$