I came across this exercise while reading a book. For context, $X$ and $Y$ are normed spaces; a linear operator $T:X\rightarrow Y$ is called compact if it maps bounded sets in $X$ to precompact sets in $Y$; a set is precompact if its closure is compact.
Of course, this result is obvious if the space $Y$ is finite-dimensional. Since the operator is bounded, any bounded set will be mapped to a bounded set and by the Heine-Borel theorem, a bounded set in a finite-dimensional space is precompact. The problem is with the infinite-dimensional spaces.
The best argument I came up with is as follows: I thought about a bounded set in $X$. Since it is bounded, it can be covered by finitely many open unit balls. Thus, it is inside the union of those open unit balls, call it $A$. Now, take $T(A)$ which will be a union of images of open unit balls, each of which is precompact. $X\subset A$, so $T(X)\subset T(A)$. Thus, $T(X)$ is precompact.
I think my argument has a problem, since I only know that the open unit ball at the origin maps to a precompact set, I don't know anything about open unit balls elsewhere. Is there anything I can do to fix my argument? Thank you.