I know this is true from various sources, unfortunately none of them give the full proof. I already have a start:
Let $G$ be connected Lie Group. Choose $K$ to be any compact neighbourhood of the identity $e$. Then the interior $\text{int}(K)$ is an open neighbourhood of $e$. Now take $V = \text{int}(K) \cap \text{int}(K)^{-1}$, then $V$ is still an open neighbourhood of $e$ and $V \subset K$.
Consider $H = \bigcup_{n \in \mathbb{N}}V^n$. This is an open subgroup of $G$. But its complement $H^C = \bigcup_{g \neq e \in G} gH$ is also open. Because $G$ is connected and $H$ is nonempty, $H = G$. In particular,
$$ G = \bigcup_{n \in \mathbb{N}}V^n = \bigcup_{n \in \mathbb{N}}K^n$$
I'm not sure how to proceed from here, though. I imagine one could take $K$ such that it is homeomorphic to a closed disk in $\mathbb{R}^n$, then this disk is second countable and thus $K$ is second countable. But even then I don't know how this translates to the entire space being second countable?
Intuitively I would assume that taking one may construct a countable dense subset of $G$ and translate the countable around the identity to each element of this subset… but I don't think that's enough?
Best Answer
The proof follows from the following lemma.
Lemma: The identity component $G_e$ of a Lie group is second countable.
Proof: Since the Lie algebra $\mathfrak{g}$ of $G$ is second countable, then so is its $k$-fold image $\exp(\mathfrak{g})\cdots \exp(\mathfrak{g})$ in $G$ for each $k ∈ \mathbb{N}$. Now $G_e$ is the countable union over $k ∈ \mathbb{N}$ of these sets.
It follows from the Lemma that a Lie group is second countable if and only if it has at most countably many components.