From the book of Munkres:
Theorem 34.1 (Urysohn metrization theorem) Every regular space $ X $ with a countable basis is metrizable.
Idea of the proof:
The hypothesis implies that there exists a countable collection of continuous functions $ f_{n}: X \rightarrow [0,1] $ having the property that given any point $ x_{0} \in X $ and any neighborhood $ U $ of $ x_{0} $, there exists an index $ N $ such that $ f_{N}(x_{0}) > 0$ and $ f_{N}(x) = 0 $ for all $ x \in X \setminus U $.
Define $ F: X \rightarrow R^{w} $ as $ F(x) = (f_{1}(x), f_{2}(x),\cdot\cdot\cdot)$.
$ F $ is an embedding (that is, $ F: X \rightarrow F(X) = [0, 1]^{w} \subset R^{w}$ is a homeomorphism).
Now $ R^{w} $ is metrizable, so the subspace $ [0,1]^{w} $ is metrizable and hence $ X $ is metrizable.
In the development of the proof of this theorem, I had the doubt of the relationship between $ [0,1] $ and $ R $ and between $ [0,1]^{n} $ and $ R^{n} $. That is,
$ [0,1]^{n} $ is homeomorphic to $ R^{n} $?
No it is not since it is compact and $\mathbb{R}^n$ is not. For the same reason $[0,1]$ is not homeomorphic to $\mathbb{R}$.