Let $K$ be a Hausdorff compact space in which every closed set is $G_{\delta}$. Suppose that $X$ is a non-metrizable subspace of $K$. Prove that $X$ cannot be written as $X=\bigcup_{n=1}^{\infty}X_{n}$ where each $X_{n}$ is a metrizable subspace of $X$.
My attempt:
It is clear that $K$ is a normal space. Furthermore, $K$ is perfectly normal. It is useful to notice that the property of being perfectly normal is hereditary.
Assume the contrary, i.e., let $X=\bigcup_{n=1}^{\infty}X_{n}$, where each $X_{n}$ is metrizable. By Bing-Nagata-Smirnov theorem, each $X_{n}$ has a countably locally finite basis. Then $X$ has a countably locally finite basis (is it true?). As noted above, the space $X$ is perfectly normal. Therefore, $X$ is regular. Consequently, $X$ is a regular space with a countably locally finite basis. Hence, $X$ is metrizable in virtue of Bing-Nagata-Smirnov theorem. This is a contradiction.
Firstly, is my proof correct? Secondly, I would like to find a proof which does not involve the metrization theorem of Bing, Nagata, Smirnov.
