Are second-countable metric spaces $\sigma$-compact?

Question

I was curious about the relations between second-countable, separable, Lindelöf and $\sigma$-compact topologies in the context of metric spaces.

I am aware of the following implications in general topological spaces:

• second-countable $\Rightarrow$ separable $\not \Rightarrow$ Lindelöf, $\;$ [thanks bof]
• $\sigma$-compact $\Rightarrow$ Lindelöf
• second-countable + locally compact $\Rightarrow$ $\sigma$-compact

as well as the reversed implications in the case of metric spaces:

• Lindelöf $\Leftrightarrow$ separable $\Leftrightarrow$ second-countable

Since all the proofs I've seen so far require the LC condition I assume it is not true in general that second-countable topological spaces are $\sigma$-compact (although seeing an actual counterexample would be nice).
So what about metrizable topological spaces?

Ideas so far:
If we can proof that every subset of a $\sigma$-compact space is again $\sigma$-compact, then this would follow from the fact, that every separable metric space is homeomorphic to a subset of the Hilbert cube (which is compact). $\;$[debunked by bof]

Answer 1

A counterexample is the "Baire space" $\mathcal{N} = \mathbb{N}^{\mathbb{N}}$. This is one of the main examples of a Polish space: a separable, completely metrizable space.

One fact about this space is that all compact subsets have empty interior, that is, all compact subsets are nowhere dense. By the Baire Category Theorem, $\mathcal{N}$ is not the countable union of nowhere dense subsets, and so together with the above fact it cannot be $\sigma$-compact.

See also this question and its answer for more details:

• Is the Baire space $\sigma$-compact?