A counterexample related to compact subset in $C[0,1]$.

Question

Define a norm on $C[0,1]$ by $||f-g||_\infty=\sup_{x\in [0,1]}|f(x)-g(x)|(<\infty)$.Now consider the metric space $(C[0,1],d)$ where $d(f,g)=||f-g||_\infty$.Consider the closed $1$-ball around $0$ i.e. $B=\{f\in C[0,1]: ||f||_\infty \leq 1\}$.We have to prove that $B$ is not compact.Consider a sequence $(f_n)$ in $B$ defined by $f_n(x)=2nx ,x\in [0,1/2n), f_n(x)=-2nx+2,x\in [1/2n,1/n), f_n(x)=0,x\geq 1/n$.Notice that $||f_n-f_m||_\infty=1$ if $m\neq n$.$\{f_n\}$ this sequence has no convergent subsequence in $C[0,1]$ because if it does,then the subsequence $\{f_{r_n}\}$ is Cauchy and $||f_{r_n}-f_{r_m}||_\infty \to 0 $ as $m,n \to 0$ which is not possible.We are done as $C[0,1]$ is not sequentially compact.

Answer 1

What you have done is correct but a much simpler example of a sequence in $B$ with no convergent subsequence is $f_n(x)=x^{n}$.