Compact operator with closed image Let $K$ be a compact operator between two normed spaces. If $K(X)$ is closed, does this necessarily imply that $K(B)$ is closed?

where $B$ is the closed unit ball?
Consider $f:c_0\rightarrow R$ defined by $f(e_i)={1\over 2^i}$, it is compact and the image of the unit ball is $(-2,2)$. where $c_0$ is the set of sequences with finitely non zero terms endowed with $\|\|_{\infty}$.
Consider $f:c_0\rightarrow R$ defined by $f(e_i)={1\over 2^i}$, it is compact and the image of the unit ball is $(-2,2)$. where $c_0$ is the set of sequences with finitely non zero terms endowed with $\|\|_{\infty}$.