Corollary 2

Every compact souslinian space is metrizable.

Let $X$ be a compact souslinian space, $I$ the interval $[0, 1]$ of $\mathbf{R}$, $H$ the set of continuous mappings of $X$ into $I$. Since $X$ is completely regular, $H$ separates the points of $X$, and cor. 1 implies the existence of a sequence $(f_n)$ of elements of $H$ which separates the points of $X$. But then the mapping $x \mapsto (f_n(x))_{n \in \mathbf{N}}$ of $X$ into $I^\mathbf{N}$, continuous and injective, is a homeomorphism of the compact space $X$ onto a subspace of $I^\mathbf{N}$; this proves that $X$ is metrizable.