For any compact topological space X, $\mathcal C(X,\Bbb R)$ is first countable? So I know that a continuous function maps compact to compact. And $\Bbb R$ is first countable. So we get that the image of X would be compact (since X is compact) and as well as first-countable due to being a subspace of $\Bbb R$ (am I right to assume that first countability is hereditary in general??). But is that what the question asks? It says "$\mathcal C(X,\Bbb R)$ is first countable?". So what does first countability of a function space mean?
Second part is a  compact subspace of $\mathcal C(X,\Bbb R)$ is closed?
I know any image of X if it is in $\Bbb R$ then compact would imply closed automatically. But again my confusion lies with what the "compact subspace of $\mathcal C(X,\Bbb R)$" refers to? 
If someone could clarify what $\mathcal C(X,\Bbb R)$ means as a space? It is a space of continuous functions from X to $\Bbb R$ right? So how does one understand countability, compactness etc for a function space? 
 A: First you have to define a topology on $C(X,\Bbb R)$. In your setting of a compact $X$, it's quite common to use the sup metric (as all continuous functions are then bounded, it's well defined):
$$d_\infty(f,g)=\sup \{|f(x)-g(x)| : x \in X\}$$
and this metric induces a topology on the function space. In this case the answer would be easy as all metric spaces $(M,d)$ are first-countable in the metric topology, as $\mathcal{U}(x)= \{B(x, \frac{1}{n}): n \in \Bbb N^+\}$ is a countable local base at $x \in M$. So that would also apply to your function space when it indeed gets the sup-metric topology.
But other topologies, like the pointwise topology, are also used and in that case it will turn out that this  function space is very often not first countable at all. So it will depend on the agreed topology on $C(X,\Bbb R)$.
A: I presume first-countable as a topological space. Anyway, I presume you are
giving $C(X,\Bbb R)$ the sup-metric. All metric spaces are first-countable, so them
$C(X,\Bbb R)$ will be.
But maybe you are giving $C(X,\Bbb R)$ the compact-open topology. But by using compactness of $X$, you can prove
that topology is the same as the sup-metric topology, so again it's first-countable.
