Prove that the closed ball $\overline{B}(0,1) \subseteq \mathcal{C}([0,1], \mathbb{R})$ can't be covered by countably-many compact sets. For $\mathcal{C}([0,1],\mathbb{R})$ the space of continuous functions $f : [0,1] \to \mathbb{R}$, I'm asked to prove that $\overline{B}(0,1) = \{ f \in \mathcal{C}([0,1], \mathbb{R}) : ||f||_\infty \leqslant 1 \}$ can't be covered by countably-many compact sets in $\mathcal{C}([0,1],\mathbb{R})$. But I'm not sure where to even begin. Any ideas on how to start?
 A: This actually holds in general for infinite-dimensional Banach Spaces, such as $\mathcal{C}([0, 1], \Bbb{R})$.
Suppose $X$ is a Banach Space. If $X$ is covered by countably many compact sets $(K_n)_{n \in \Bbb{N}}$, then because each set is closed, the Baire Category Theorem implies that at least one of the $K_i$s has a non-empty interior. That is, there exists some closed ball $x + rB_X \subseteq K_i$, where $r$ is some positive real number.
But then, this implies,
$$B_X \subseteq \frac{1}{r}(-x + K_i),$$
which is compact, hence $B_X$ is compact. A compact closed unit ball implies that $X$ is finite-dimensional. Thus, if $X$ is infinite-dimensional, no such compact sets can exist.
A: Suppose for each $n\in \mathbb{N}$, $C_n$ is compact. Then the $C_n$'s are equicontinuous; choose the $\delta_n$ as in the definition of equicontinuity of $C_n$. It is easy to construct a continuous function $f:[0,1] \rightarrow [-1, 1]$ whose graph has gradient greater than $\frac{1}{\delta_n}$ at $1/n$; the graph merely has to consist of countably many tents next to each other of required steepness. To make $f$ continuous just require that the tent width tends to $0$. Then $f\notin \bigcup\limits_n C_n$.
