Show that the closed unit ball $B[0,1]$ in $C[0,1]$ is not compact under the following metrics:
$1. d(f,g)=\sup_{x\in [0,1]}|f(x)-g(x)|$
$2.d(f,g)=\int _0^1 |f(x)-g(x)| dx$
My try:
In order to show not compact if we can find a sequence which has no convergent subsequence then we are done.
How to approach $1,2$?
On
You haven't provided a set of functions for the first sequence. What are your $e_n$'s ?
For the first sequence, consider the set of functions $p_n(x) = x^n$. We have $\sup_{[0,1]} |x^n| = 1$ for all $n$, since $1^n = 1$. However, no subsequence converges uniformly to a continuous function. They converge to the function $$f(x) = \left\{ \begin{array}{cc} 0 & x \neq 1\\ 1 & x=1\end{array}\right.$$
For the second part, I suggest following up on @GregoryGrant's post.
On
The uniform case is classic: use $f_n(x)=x^n$. There is a lot to be learned from this example.
That example won't work in the integral case, because this sequence converges to the zero function in the integral sense. Instead, try making a function which is a continuous approximation to a function with a jump: for instance it is $1$ on $[1/4,3/4]$, zero outside $[1/4-1/(4n),3/4+1/(4n)]$, and continuous (say, linear) in between. These look like trapezoids which are approaching a rectangle.
Another possibility is $\sin(nx)$, but I think the proof might be more difficult in this case.
On
For 1. take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}),1),({1 \over n}, 0), (0,0) $. Then $d(0,f_n) = 1$, and $d(f_n,f_m) = 1$ for all $n \neq m$ hence there are no convergent subsequences.
$f_n$ is a 'tooth' of height $1$ with support in $[{1 \over n+1},{1 \over n}]$.
For 2. in a similar fashion, take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}), 2n (n+1)), ({1 \over n}, 0)), (0,0) $. Then , as above, $d(0,f_n) = 1$, and $d(f_n,f_m) = 2$ for all $n \neq m$ hence there are no convergent subsequences.
$f_n$ is a 'tooth' of height $2n(n+1)$ with support in $[{1 \over n+1},{1 \over n}]$.
On
As mentioned in previous answers, an animation much aids comprehension. Here's an open cover for the closed unit ball with respect to sup norm metric. Let,
\begin{align} f^n_{\mathrm{upper}}(x) &= \text{the continuous function obtained by connecting the points } \begin{cases} (0,0),\\ (1/n,2),\\(1-1/n,2),\\(1,0) \end{cases}\\ f^n_{\mathrm{lower}}(x) &= \text{the continuous function obtained by connecting the points } \begin{cases} (0,0),\\ (1/n,-2),\\(1-1/n,-2),\\(1,0) \end{cases}. \\ \end{align} And then define each open set, $$ \mathcal{O}_n = \{ g(x) \in C[0,1] : f^n_{\mathrm{lower}}(x) < g(x) < f^n_{\mathrm{upper}}(x) \quad \forall x \in [0,1]\}. $$
And you can check that every function in the unit closed ball lies in at least on of these $ \mathcal{O}_i$s, and each $ \mathcal{O}_i$ is an open set. From the illustation it's clear that the cover does not contain a finite subcover.
The illustration is created using Desmos.
Use functions that are $n^2$ on $[0,1/n]$ and zero on $[1/n+\epsilon,1]$ and on $[1/n,1/n+\epsilon]$ connect them to be continuous. Then the integral of any one of them is close to $n$ (choose an appropriate $\epsilon$ that will depend on $n$) and I believe the integral of the difference of any two of them is at least one.