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

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]

• The set of irrational numbers is a separable metric space which is not $\sigma$-compact. (Every $\sigma$-compact subset of the irrational numbers is meager.) Hilbert space is another example. – bof Nov 12 '18 at 11:10
• By the way, separable does not imply Lindelöf in general topological spaces. For example, the Sorgenfrey plane is separable but not Lindelöf. – bof Nov 12 '18 at 11:12
• Every closed subset of a $\sigma$-compact space is again $\sigma$-compact. Not all subspace, as witnessed by the rational and the irrationals assubspaces of the reals. – Henno Brandsma Nov 12 '18 at 11:55
• @bof I also wanted to disprove local compactness. – Henno Brandsma Nov 12 '18 at 12:25

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.