A sufficient and necessary condition for a special linear operator to be compact

Here is an idea: let $T(X) \ni y_n=T(x_n)\to y\in \overline{T(X)}$. Then, since $\|Tx_n\| \ge c\|x_n\|,\ x_n\to x\in X$ and continuity of $T$ implies that $y_n=T(x_n)\to T(x)$ and so $y=T(x)$. Therefore, $T$ has closed range.

$T(B_X)$ contains a ball in the Banach space $T(X)$, by the open mapping theorem, and since $\overline {T(B_X)}$ is compact, $T(X)$ is locally compact, hence finite dimensional. But $T$ is injective, so $X$ must be finite dimensional ,too.