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?

• Take the linear functional $x=(1/2,1/4,1/8,\ldots)\in\ell_1$ on $X=c_0$. Its range is all of $\Bbb R$, but the image of $B(c_0)$ under $x$ is $(-1,1)$. – David Mitra Jul 23 '16 at 18:46

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 $\|\|_{\infty}$.
• So $(-2,2)$ is closed but not compact, right? – MorganeMaPh Jul 23 '16 at 18:48
• But in your example, is $f(e_i)$ closed? – MorganeMaPh Jul 23 '16 at 18:53