Given $X = (C[0,1], d_\infty)$ and $f_n \in X$. I want to show that for every compact set $K \subseteq X$, $\{f_n : n \geq 1\} \nsubseteq K$.
Where $f_n(x) = \begin{cases} 0, &\quad 0 \leq x \leq \frac{1}{n+1} \\ \frac{1}{\frac{1}{n+1/2} - \frac{1}{n+1}}\cdot (x-\frac{1}{n+1}), &\quad \frac{1}{n+1} \leq x \leq \frac{1}{n+1/2} \\ \frac{1}{\frac{1}{n+1/2}-1/n}\cdot(x-1/n), &\quad \frac{1}{n+1/2} \leq x \leq 1/n\\ 0, &\quad 1/n \leq x \leq 1 \\ \end{cases} $
I am having a hard time proving this, since I haven't seen any exercises alike. I thought maybe I could use the fact that if $K$ is compact and $f_n(x) \subseteq K$ then it must have a convergent subsequence. Would the idea be then to show that the sequence of functions $f_n$ has no convergent subsequence in $C([0,1], d_\infty)$?