By Bolzano-Weierstrass, any bounded sequence $(x_k)$ in $\mathbb R^n$ has a convergent subsequence. Now try to construct a sequence sequence that does not converge, assuming a set is closed and bounded, yet there exists an open covering that has no finite subcover.
Let $A \subset \mathbb R^n$ be closed and bounded, s.t. there exists an open covering which has no finite subcovering. So let $U_i$ be such an open covering. Pick an open set, which we denote by $U_0$. Let
$x_0 \in A - U_0$ , which exists by assumption.
Choose $U_1 \in (U_i)$ with $x_0 \in U_1$. This exists by assumption.
Now we proceed in that manner and choose a sequence of open sets from the covering and a sequence in $A$ that is always one step before the covering sequence.
Then the constructed sequence $(x_k)$ is bounded, as it is contained in $A$. It has a convergent subsequence, say, to $x$. Choose such a sequence and keep the label $x_k$. As $A$ is closed, $x \in A$.
By assumption, we can choose some $V$ which covers $x$. We do so at any point of the sequence, say after some fixed number of steps. We strip off all elements of the sequence which are in $V$, and construct the two sequences again in the same fashion, with $V$ being included in the sequence of open sets that we used to define the sequence $x_k$.
We repeat this. Thus we construct a bounded sequence that does not converge - because otherwise, it would be covered by an open set in the sequence we have choosen.
Hence, if $A$ is closed and bounded, every open covering has a finite subcover. $qed$