Suppose that $X=\bigcup\_{n=1}^\infty K\_n$ is a topological space, $K\_n$ is a metrizable subspace in $X$ for every $n \in \omega$, then $X$ is a metrizable space?


In metrizable spaces, compactness is equivalent to $\sigma$-compactness?


**One more:** Is pseudocompactness hereditary with respect to $\sigma$-compact subspaces?



Are countable unions of metrizable spaces metrizable too?
No to all your questions. There are lots of countable non-metrizable spaces: an easy one is $\mathbb{N}$ in the cofinite topology, which can be written as a countable (disjoint) union of singletons (which are metrizable, and closed and compact). Using ultrafilter spaces (given an ultrafilter $\mathcal{F}$ on $\mathbb{N}$, define $X = \mathbb{N} \cup \{\infty\}$ where $\mathbb{N}$ is discrete and a neighbourhood of $\infty$ is of the form $A \cup \{\infty\}$, where $A \in \mathcal{F}$); such spaces are countable, hereditarily normal but not metrizable (not even first countable at $\infty$). 


All these spaces above are $\sigma$-compact but not compact, and of course a discrete countable set like $\mathbb{N}$ is metrizable, $\sigma$-compact but not compact, as is $\mathbb{R}$, e.g. 


Also $[0,1]$ is pseudocompact, but the $\sigma$-compact subset $\{ \frac{1}{n} \mid n \in \mathbb{N} \}$ is not (pseudo)-compact.


