Let $\langle X,d\rangle$ be a non-separable metric space. If $A\subseteq X$ is countable, then $\operatorname{cl}A\ne X$, so we can choose a point $x\in X\setminus A$. This means that we can recursively construct a set $\{x_\xi:\xi<\omega_1\}\subseteq X$ such that if $F_\eta=\operatorname{cl}\{x_\xi:\xi<\eta\}$ for each $\eta<\omega_1$, then $x_\xi\notin F_\xi$ for $\xi<\omega_1$. For each $\xi<\omega_1$ there is an $n(\xi)\in\omega$ such that $B\left(x_\xi,2^{-n(\xi)}\right)\cap F_\xi=\varnothing$. For $k\in\omega$ let $A_k=\{\xi<\omega_1:n(\xi)=k\}$; there is some $m\in\omega$ such that $A_m$ is uncountable. Show that $\left\{B\left(x_\xi,2^{-(m+1)}\right):\xi\in A_m\right\}$ is an uncountable family of pairwise disjoint open sets. Conclude that if a metrizable space has the Suslin property, it must be separable. (The other direction is trivial.)