(This is just an expansion of my comment above.)

As Buschi Sergio points out, a topological space that is the union of countably many metrizable subspaces need not even be Hausdorff. My example of CW-complexes is intended to show that even when such a space is Hausdorff (and paracompact and submetrizable), it can still easily fail to be metrizable.

Every CW-complex is $\operatorname{F}_\sigma$-metrizable (that is, the union of countably many closed metrizable subsets). This is easy to see when there are only countably many cells, as the cell closures are compact and metrizable. But it's true in general, as I'll explain below. (I should perhaps clarify that by a "cell" of a CW-complex I mean what is sometimes called an "open cell". The cells partition the space.)

If a subset $A$ of a CW-complex is such that $A\cap e$ is compact for each cell $e$, then $A$ is closed and metrizable. (It's metrizable because it's the topological sum of the compact sets, which are metrizable.) Since each cell is $\sigma$-compact, we can express the CW-complex as a union of countably many closed metrizable subspaces of this form.

The simplest example of a CW-complex that is not metrizable consists of a single 0-cell and a countable infinity of 1-cells, forming a bouquet of circles. This is not metrizable as first-countability fails at the 0-cell.

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