Is image of ball of finite rank linear operator compact?

If $X$ is not reflexive, then a counterexample can be formed using James' theorem. Since $X$ is not reflexive, there must exist a functional $f \in X^*$ such that $f$ fails to achieve its maximum on the closed unit ball. Take such a functional, and $f(B_X)$ becomes the open ball of $\mathbb{C}$.

Let $X=c_0$, the space of complex sequences that converge to 0 with the sup norm. Define $f:X \to \mathbb C$ by $f(c_n)=\sum \frac {c_n} {n^{2}}$ Then the image of the unit ball of $c_0$ is not closed because $|\sum \frac {c_n} {n^{2}}| \leq \sum \frac 1 {n^{2}}$ for all $(c_n)$ in this ball and $\sup \{ |f(c_n)|:||(c_n)|| \leq 1\} = \sum \frac 1 {n^{2}}$ but the value $\sum \frac 1 {n^{2}}$ is not attained (since we cannot have $|c_n| =1$ for all $n$).