Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?

The results requires some choice. It is consistent that $\mathbb R$ is not Lindelöf but still second countable (without the axiom of choice).

The axiom of choice for countable families of subsets of $\mathbb R$ is in fact equivalent that second countable implies Lindelöf, and to a few other interesting assertions:

H. Herrlich and G. E. Strecker, When is $\mathbb{N}$ Lindelöf?, Comment. Math. Univ. Carolinae 38,3 (1997), 553-556.

I think it's worth pointing out that, although second-countability implies Lindelöf requires some choice, the Heine-Borel theorem itself doesn't. To prove that a closed and bounded subset of $\mathbb{R}$ is compact, consider first the case $[0,1]$.

Given an open cover of $[0,1]$, consider the set $S = \{x \in [0,1]\mid\text{there is a finite subcover of }[0,x]\}$. Then $0 \in S$, $S$ is open, and the least upper bound of $S$ is an element of $S$ (check each of these facts). It follows that $1 \in S$ is the least upper bound, so there is a finite subcover.

To prove that any closed and bounded set is compact, realize that it is a closed subset of $[-a,a] \cong [0,1]$ for sufficiently large $a$.