**Step 1: Forward direction ($\Rightarrow$).** Suppose $V$ is $n$-dimensional with basis $\{e_1, \ldots, e_n\}$. By the [Equivalence of Norms on Finite-Dimensional Spaces](/theorems/877), all norms on $V$ are equivalent, and compactness is preserved under Lipschitz equivalence of norms (because a convergent subsequence in one norm is convergent in any equivalent norm). Hence it suffices to prove compactness of $\overline{B}_1(0)$ in the $\ell^1_n$ norm $\|v\|_{\ell^1_n} = \sum_{i=1}^n |v_i|$. This was established in the proof of the [Equivalence of Norms](/theorems/877): every [sequence](/page/Sequence) in $\overline{B}_1(0)$ has componentwise bounded coordinates, and iterated application of Bolzano-Weierstrass produces a convergent subsequence. **Step 2: Reverse direction ($\Leftarrow$).** Suppose $V$ is infinite-dimensional. We construct a sequence in $\overline{B}_1(0)$ with no convergent subsequence. [claim:Covering Argument] $\overline{B}_1(0)$ cannot be covered by finitely many open balls of radius $1/2$. [/claim] [proof] Consider the open cover $\overline{B}_1(0) \subset \bigcup_{y \in \overline{B}_1(0)} B_{1/2}(y)$. If $\overline{B}_1(0)$ were compact, there would exist a finite subcover: $\overline{B}_1(0) \subset \bigcup_{i=1}^m B_{1/2}(y_i)$. Set $Y = \operatorname{span}\{y_1, \ldots, y_m\}$, which is a finite-dimensional proper subspace of $V$. By the covering, every $v \in \overline{B}_1(0)$ satisfies $v \in Y + B_{1/2}(0)$. Rescaling: $B_{1/2}(0) \subset Y + B_{1/4}(0)$ (since $Y/2 = Y$). Iterating: $\overline{B}_1(0) \subset Y + B_{1/2^k}(0)$ for all $k \ge 1$. The intersection $\bigcap_{k=1}^\infty (Y + B_{1/2^k}(0))$ equals $\overline{Y}$. To verify: if $v \in Y + B_{1/2^k}(0)$ for all $k$, then for each $k$ there exist $u_k \in Y$ and $w_k$ with $\|w_k\| < 1/2^k$ such that $v = u_k + w_k$. Hence $\|v - u_k\| < 1/2^k$, so $u_k \to v$. Since $Y$ is finite-dimensional, it is complete (by the [Equivalence of Norms](/theorems/877) and the completeness of $\mathbb{R}^n$), hence closed, so $v \in Y$. Therefore $\overline{B}_1(0) \subset Y$. But $Y$ is finite-dimensional and $V$ is not, so there exists $v \in V \setminus Y$. For $R > 0$ large enough, $v/R \in \overline{B}_1(0) \subset Y$, contradicting $v \notin Y$ (since $Y$ is a subspace). This contradiction shows $\overline{B}_1(0)$ is not compact. [/proof]