Suppose $f_i: X \rightarrow \mathbb{R}$, $i \in I$ where $Χ$ is second countable and $f_i$ are continuous $\forall i \in I$. Then the family $(f_i)_{i \in I}$ is separating the points of $X$?
If $X$ was normal space we could apply Urysohn's Lemma, but here $X$ is second countable (and consequently separable).
Any ideas would be helpful
For any space, if the $f_i$ happen to be all constant, or on $X=\mathbb{R}$ we take the functions $f(x) = A\sin(x), A \in \mathbb{R}$, say, we cannot separate all points of $X$. So you have to be careful about the family of functions.
There are demands on $X$ as well:
If $X$ is indiscrete (e.g.) then $X$ is second countable, and all continuous functions on $X$ with codomain $\mathbb{R}$ (or any Hausdorff space) are constant.
So no family of continuous functions on $X$ can separate the points.
$X$ is called functionally Hausdorff iff such a point-separating realvalued-continuous family of $X$ exists at all. Not all spaces are functionally Hausdorff, as we saw. Even $T_3$ spaces need not be.
A second countable regular $T_1$ space (which is normal so Urysohn applies) will have a (countable, even) family of this type.