Image of closed unit ball under a compact operator

It is false in general that the image of the closed unit ball under a compact operator is closed (and hence compact). Here is an easy example:

Consider $X = C[0,1]$ with the uniform norm, and the compact operator $A \in B(X)$ defined by the formula:

$\displaystyle\qquad Af(x) = \int_0^x f(t)\,dt$.

Compactness of $A$ is easily proven using Arzelà–Ascoli. Our operator $A$ produces an anti-derivative of any input given to it, and the image of the closed unit ball of $X$ under $A$ is the set

$\displaystyle\qquad \{f \in C^1[0,1] \mathrel: f(0)=0,\ \lVert f'\rVert \leq 1\}$

which certainly is not a closed subset of $X$.