Compact operators in finite dimensional spaces Let $X$ and $Y$ be finite-dimensional Banach spaces. Assume $f\colon X\to Y$ is continuous. Take any bounded subset $M$ of $X$. Then $cl\,M$ is compact. Since $f$ is continuous, we deduce that $f(cl\, M)$ is also compact. From the properties of continuous functions I know that
$$f(cl\, M)\subset cl\,f(M).\qquad (1)$$
To show that $f$ is compact operator, I need "=" instead of "$\subset$" in (1). Can we deduce it somehow?

You can say that for $A\subset X$ bounded, then $f(A)\subset Y$ is bounded because of continuity of $f$. Moreover, since we are in a metric space, the closure of a bounded set is bounded (try to prove by contradiction), so that $cl(f(A))\subset Y$ is a closed bounded set, now $Y$ is linearly isomorphic to $\mathbb{R}^d$ for some $d$ since it is finite dimensional (general result of linear algebra), therefore you conclude that closed bounded iff compact, that is, $cl(f(A))$ is compact.
