Let $X$ be a set of continuous function from the space $\mathbb R$ of real with the absolute value topology into itself. For each $f \in X$ and $p>0$ define
$N_p(f)=\{g\in X: |f(x)-g(x)|<p, \forall x \in \mathbb R\}$
the family of $N_p(f)$ for each $f \in X$ and $p>0$ form the basis for a topology $\tau$ on $X$. Is $X$ second countable?
Let $Y=\{f\in X: f$ has derivatives of all orders at each $x\in X \}$
Is $Y$ a second countable subspace of $X$?
From the second part of the question I guess that $X$ must be second countable, othewise they will not ask stupid question like that. However, I know that the family of $N_p(f)$ for each $f \in X$ and $p>0$ form the basis for a topology $\tau$ on $X$, but I don't see any sign showing that this basis is countable.
$X$ is not separable. For every subset $A \subset \mathbb{Z}$ we can make a function $f_A$ that is $0$ everywhere except $1$ at each $n \in A$, with a small peak up from the base level of $0$ to make it continuous. Then for each $A \neq B$ the distance between $f_A$ and $f_B$ equals $1$, so all open balls around different $f_A$ (of which there are continuum many) of radius $\frac{1}{3}$ are all mutually disjoint and any dense set should intersect them all. And a space that is second countable has all subspace separable...
[Added] $Y$ is similarly non-separable, because we can make the peaks infinitely differentiable if we want.