= Solution
Two <norms> are <Lipschitz equivalent norms> if constants $c,C>0$ satisfy $c\|u\|_1\leq\|u\|_2\leq C\|u\|_1$ for every $u$. On the <vector space> of finitely supported sequences, the <norms> $\|x\|_1=\sum|x_i|$ and $\|x\|_\infty=\max|x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two <norms> $N$ and one.
For dimension $d<\infty$, choose a <basis> and write $u=\sum x_ie_i$. Any <norm> satisfies $\|u\|\leq\sum|x_i|\|e_i\|\leq C|x|_2$. This also implies its <continuity> in the Euclidean coordinates, by $|\|u\|-\|v\||\leq\|u-v\|\leq C|x-y|_2$. Its positive continuous values on the <compact> Euclidean <unit sphere> have a positive minimum $c$. Homogeneity gives $c|x|_2\leq\|u\|$, proving equivalence with the Euclidean <norm> and hence between any two <norms>.
If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient <norm>, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in $\mathbb R^d$ or $\mathbb C^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus \b[finite-dimensional subspaces are closed].
A finite-dimensional normed space has nonempty open balls with <compact> closures by equivalence and Heine-Borel. Conversely let an <open set> $O$ have <compact closure>. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a <closed subset> of the <compact closure> of $O$ and is therefore <compact>. Translation and scaling make the closed <unit ball> <compact>.
If the space were infinite-dimensional, construct unit vectors $x_j$ with $\operatorname{dist}(x_j,\operatorname{span}(x_1,\ldots,x_{j-1}))>1/2$. Here is the needed <Riesz lemma>: for a proper closed subspace $M$, choose $z\notin M$, put $d=\operatorname{dist}(z,M)>0$, and choose $m\in M$ with $\|z-m\|<2d$; then $(z-m)/\|z-m\|$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting <compactness>. Therefore \b[the space is finite-dimensional exactly when it has a nonempty <open set> with <compact closure>].
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