Let $X$ be a Banach space and $A\subset X$ be bounded. Suppose that, for any $\varepsilon>0$, $\exists\; F_\varepsilon$ a subspace of $X$ of finite dimension such that $\text{dist}(x,F_\varepsilon)\leq\varepsilon,\quad\forall x\in A.$
Then prove that $A$ is relatively compact in $X$, i.e. its closure is compact in $X$.
Can anybody give me an hint?