Prove that a separable metric space is Lindelöf without proving it is second-countable This a problem from my exam:

Prove that a separable metric space is a Lindelöf space

I know that a separable metric space is second-countable, and second-countable implies the Lindelöf property. But I wonder whether it is possible to prove directly that "a separable  metric space is a Lindelöf space" without proving it to be second-countable, because it's cumbersome to prove this by proving 2 other not easy lemmas.
In this way, I find it difficult to find countable subcover for a given open cover. Can any one help me? I really appreciate.
 A: Let $\mathcal{U}$ be an open cover of $(X,d)$. Let $\{d_n: n \in \mathbb{N}\}$ be a dense subset of $X$. For every $n\in \mathbb{N}, q \in \mathbb{Q}$, if there exists some member $U$ of $\mathcal{U}$ that contains $B(d_n, q)$, pick some $U(n,q) \in \mathcal{U}$ that does (there could be plenty of such $U$ so we use AC to pick definite ones). Otherwise $U(n,q) = U_0$ for some fixed $U_0 \in \mathcal{U}$, for definiteness.
Claim: $\{ U(n,q): n \in \mathbb{N}, q \in \mathbb{Q}\}$ is a countable subcover of $\mathcal{U}$. 
To see this, let $x \in X$ and find some $U_x \in \mathcal{U}$ that contains $x$ (as we have a cover). Then for some $r>0$ we have $B(x,r) \subset U_x$, and we can also find some $d_{n(x)}$ from the dense subset inside $B(x,\frac{r}{2})$, and next we find a $q(x) \in \mathbb{Q}$ such that $d(x, d_{n(x)}) < q(x) < \frac{r}{2}$. 
Note that $x \in B(d_{n(x)}, q(x))$ and $B(d_{n(x)}, q(x)) \subset B(x,r) (\subset U_x)$ by the triangle inequality.
Then $U_x$ witnesses that some member of $\mathcal{U}$ contains $B(d_{n(x)},q)$ and so we know that $x \in B(d_{n(x)}, q(x)) \subset U(n(x), q(x))$ and so $x$ is indeed covered by the subcover, as required.
This proof is quite direct, but it is essentially the same argument one needs to see that all sets $B(d_n, q)$ for $n \in \mathbb{N}, q \in \mathbb{Q}$ form a (countable) base for $X$. So it's not really a simplification per se, but just to show a direct proof is possible.
BTW, it's not too hard to go to hereditarily Lindelöf as well. Essentially the same proof holds: $\mathcal{U}$ is then not necessarily a cover of $X$ and the $x$ is then chosen to be in $\cup \mathcal{U}$ instead of $X$. We use the formulation of hereditarily Lindelöf as : every family of open sets has a countable subfamily with the same union. The proof goes through the same way, really.
