understanding the proof that a compact operator in infinite dimensional space is not surjective

Question

Let $X, Y$ be infinite-dimensional Banach spaces. If $K : X \to Y$ is a compact linear operator, prove that $K(X)\neq Y$ , i.e., $K$ cannot be surjective.

I have problem to understand the following proof. I bolded the part

Assume that $K$ is surjective. By the open mapping theorem, $K$ is an open map. In particular, the image of the unit ball $B_1:= \{x \in X : \|x\| < 1\}$ contains a neighborhood of the origin. $\textbf{But this is impossible, because the closure $K(B_1)$ is compact, and cannot contain any open set}$ $\textbf{in the infite dimensional space Y}$ .

Answer 1

If $K(B_1)$ is compact and contains $B(0,r)$ its closure contains the closed ball $\bar B(0,r)$ which is not compact since $Y$ is infinite dimensional, contradiction since a closed subset of a compact set is compact.

Is it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?