Suppose $X$ is Banach and $T\in B(X)$ (i.e. $T$ is a linear and continuous map and $T:X \to X$). Also, suppose $\exists c > 0$, s.t. $\|Tx\| \ge c\|x\|, \forall x\in X$. Prove $T$ is a compact operator if and only if $X$ is finite dimensional.
"$X$ is finite dimensional $\implies$ $T$ is compact" is easy to show. To prove the other side, at first, I made a mistake, thinking $X$ is reflexive. Then this work can be easily done by the fact that any sequence of a reflexive linear space has a weakly convergent subsequence and $T$ is completely continuous (since $T$ is compact). But this is not the situation.
So how to prove "$T$ is compact $\implies X$ is finite dimensional"?