Let a metric space $(X,d)$ be separable. Then $|X|\le \mathfrak{c}$.
The way I'm thinking of proving this is as follows:
(1) Since $(X,d)$ is separable, there is a dense subspace $Y\subset X$ such that $\overline{Y}=X$, with $Y$ being countable.
(2) Then there exists an injection $f:Y\to \mathbb{N}$.
(3) But $|\mathbb{N}|=|\mathbb{Q}|=\aleph_0$. Thus there exists a bijection $g:\mathbb{N}\to\mathbb{Q}$.
(4) Then $g\circ f:Y\to \mathbb{Q}$ is an injection.
(5) But $\mathbb{R}$ can be constructed in the following way:
$$\mathbb{R}=\{(q_n)_{n=1}^\infty : \forall 0<\varepsilon\in\mathbb{Q}, \exists n_\varepsilon\in\mathbb{N}\mbox{ such that } m,n\ge n_\varepsilon \implies |q_m-q_n|<\varepsilon \}$$
(6) So there is an injection $h: \mathbb{Q}\to \mathbb{R}$.
(7) Thus $h\circ g\circ f: Y\to \mathbb{R}$ is an injection.
(8) Hence, $|X|\le \mathfrak{c}$.
Please let me know whether or not you find this proof outline to be correct.
Your proof is not correct, all you show is that there is an injection from a countable set into the real numbers. And well... isn't that obvious since $\Bbb{N\subseteq R}$?
Here is a key step in the proof: In a metric space, sequences determine the topology. Specifically, every point in the metric space is the limit of a sequence from $Y$, your countable dense subset.
How many sequences are there?
(For bonus points, show that we can actually find a canonical sequence converging to each point, after fixing an enumeration of $Y$.)