Let $T: \mathcal{B_1} \rightarrow \mathcal{B_2}$ be linear and onto such that $T(B_{\mathcal{B_1}}(0,r))$ is contained in a compact, for each $r > 0$. Show that $dim \mathcal{B_2} < \infty$.
My attempt: I want to show that $\overline{B_{\mathcal{B_2}}(0,1)}$ is not compact in $\mathcal{B_2}$, because we have that $\overline{B_{\mathcal{B_2}}(0,1)}$ if, and only if, $dim \mathcal{B_2} < \infty$.
Suppose that $B_1$ and $B_2$ are Banach. Then the open mapping theorem implies that $T$ is open since it is surjective. This implies that the image of $B_{B_1}(0,r)$ contains an open subset, thus it contains also an open ball $B_{B_2}(0,r')$ which is contained in $T(B_{B_1}(0,r))$. This implies that the adherence of $B_{B_2}(0,r')$ is compact since $B_{B_2}(0,r')\subset T(B_{B_1}(0,r))\subset C$ where $C$ is compact. We deduce that $B_2$ is finite dimensional since a Banach space for which closed balls are compact is finite dimensional.