Show that if a metric space is complete, separable and not countable then it has cardinal $\aleph_1$ 
Show that if a metric space is complete, separable and not countable then it has cardinal $\aleph_1$

I have encountered this exercise and I don't know where to start. There is a lot of important information and I can't imagine how to use it. Any hint?
 A: Although the answer @Shalop contributed with is surely correct, this course is about metric spaces, not topological spaces. And that's why when dealing with the concepts of given a topology, etc, I can't follow. However, I came up with a proof, that in a similar manner, I don't see why $X$ has to be complete.
Since $X$ is separable, it contains a countable dense subset $\mathcal{D} \subset X$. We have that $\overline{\mathcal{D}}=\mathcal{D} \cup \{\text{the set of limit points of } \mathcal{D}\}=X$. (Here $\overline{\mathcal{D}}$ stands for the closure of $\mathcal{D}$). 
Now, since $\mathcal{D}$ is countable, then the "number" of sequences in $\mathcal{D}$ is $\aleph_{0}^{\aleph_{0}}=\aleph_{1}$ Therefore, $\#  \{\text{the set of limit points of } \mathcal{D}\} \leq \aleph_{1}$. 
However $\#  \{\text{the set of limit points of } \mathcal{D}\} > \aleph_{0}$. Because if not $\# (\overline{\mathcal{D}})=\#(\mathcal{D} \cup \{\text{the set of limit points of } \mathcal{D}\})= \# (X) = \aleph_0$ Which contradicts the fact that $X$ is not countable. 
Therefore $\# (\overline{\mathcal{D}})= \# (X) = \aleph_{1}$
A: We will show that the cardinality of any second-countable $T_1$ space is at most $\aleph_1$, which includes this as a special case.
Let $X$ be such a space, and let $\{U_n \}_{n \in \mathbb{N}}$ be a countable basis for $X$.
Let $\tau$ denote the topology (i.e, the collection of open sets) of $X$, and define a map $f: \mathcal{P}(\mathbb{N}) \to \tau$ as $J \mapsto \bigcup_{k\in J} U_k$. Then $f$ is surjective since $\{U_k\}$ forms a basis for $\tau$. Therefore $|\tau| \leq |\mathcal{P}(\mathbb{N})| = \aleph_1$.
Next, define a map $g: X \to \tau$ by $x \mapsto X \setminus \{x\}$. Then $g$ is injective, and thus $|X| \leq |\tau| \leq \aleph_1$.
Edit: It has been pointed out to me that the above argument only shows that $|X| \leq \aleph_1$, not that $|X| = \aleph_1$. Assuming the continuum hypothesis, this would automatically hold (given that $|X|> \aleph_0$), and we would be done.
If we don't assume CH, we can still obtain the result that $|X| \geq \aleph_1$ using complete metrizability of $X$. I don't see a truly elementary argument for this, but see theorem A.4 in the following set of notes for a terse little proof: http://www.math.tamu.edu/~kerr/book/appendixA.pdf
A: Joaquin's and Shalop's answers prove the desired result without assuming completeness but using the continuum hypothesis (CH). In fact, they show that $\#X\leq\#\mathbb R$, and for this particular result there is no reliance on the CH.
Here is a somewhat involved proof (Lemma 2) that every uncountable Polish space (that is, separable and complete metric space) has cardinality of at least that of $\#\mathbb R$; this proof does exploit completeness. Hence, it is possible to give a complete solution to Joaquin's question without relying on the CH.
