I don't know of any result of the form
Theorem: Let $(V,\|\cdot\|)$ be [$\ldots$]. Then $V$ is finite / infinite dimensional.
Proof: let us show that the unit ball is compact / non-compact. $[\ldots]$
At least, I don't know of any such results where no simpler proof exists.
However, the equivalence
$(V,\|\cdot\|)$ is finite dimensional $\iff$ the unit ball is compact.
is of crucial theoretical importance.
Note that the fact that the unit ball is compact is equivalent to the fact that closed bounded domains are compact, or to the fact that bounded sequences admit limit points.
Thus, the above result tells you that one cannot expect to use compactness arguments in infinite dimensional normed vector spaces to show existence of solutions to some minimising problems or some PDEs.
Below is a hand-wavy illustration.
Assume $(V,\|\cdot\|)$ is finite dimensional, and let $F \colon V \to [0,+\infty)$ be some continuous functional.
Assume that $F(u) \to \infty$ whenever $\|u\| \to \infty$.
Then there exists $u_0 \in V$ such that $F(u_0) = \min_{u\in V} F(v)$.
Indeed, one easily builds a sequence $u_k$ such that $F(u_k) \leqslant \inf F + \frac{1}{k}$, then shows that $\{u_k\}$ must be bounded, and exhibits a converging subsequence thanks to a compactness argument.
The limit $u_0$ then solves the problem.
Now, assume that $(V,\|\cdot\|)$ is not finite dimensional.
Then one can still create a "minimizing" subsequence $\{u_k\}$ such that $F(u_k) \to \inf F$.
However, absolutely nothing can ensure the existence of a converging subsequence, and hence, you cannot expect to exhibit a solution to the problem this way.
This example is not artificial: some PDEs are naturally the Euler-Lagrange equations of some suitable functionals as above, and solving those said PDEs boils down to finding minimizers of the functionals.
Since function spaces, such as $\mathcal{C}^{\infty}(X)$, $\mathcal{C}^{k,\alpha}(X)$, $\Bbb C[X]$, or $H^p(X)$, are infinite dimensional, you can't expect to use a normed topology to solve many problems in functional analysis, at least not in this way.
One has to build appropriate tools to make the above heuristic work.
Some of these tools are completeness, compact embeddings of Sobolev spaces, etc.
One can also abandon the normed topology and look for a "better" topology that reflects the properties of the problem, such as weak topology, weak-* topology, and so on.
Best Answer
Indeed, let $(F, \lvert \cdot \rvert)$ be any field with an absolute value, and $(V, \|\cdot\|)$ be a normed $K$-vector space with respect to the that absolute value (whether that norm and/or absolute value satisfy the strong triangle inequality or not). Then w.r.t. the metric induced by the norm, we have that
Because as you suggest, using that for any $v_0 \in V$ the translation map $x \mapsto v_0+x$ is a homeomorphism (which can be proved right away, or follows just from the fact that the underlying additive group of $V$ is a topological group w.r.t. the considered structure), we get that $v_0 + \bar B_1(0)$ is a compact neighbourhood of $v_0$.
For the converse direction
we need only one little extra restriction, namely we have to assume that the absolute value $\lvert \cdot \vert$ is not trivial, i.e. there is some $c \in F$ such that $\lvert c \rvert \neq 0,1$.
Namely, even without that assumption, from local compactness we can readily find some closed ball of radius $r > 0$, $\bar B_{r}(0) :=\{v \in V: \| v \| \le r\}$ around $0 \in V$ which is compact; now we need that assumption to further say w.l.o.g. that $r = \lvert a^{-1} \rvert$ for some scalar $a \in F$; and again, quite generally scaling $x \mapsto ax$ is a homeomorphism for any $a \in F^\times$, and will map that closed ball into and onto the closed unit ball.
(And in that trivial case that the absolute value and hence the norm is trivial, say $\lvert v \rvert = \begin{cases} c \text{ if } v \neq 0 \\ 0 \text{ if } v=0 \end{cases}$ for some $c > 0$, just notice that such a $V$ is always locally compact, whereas closed balls $\bar B_{r}(0)$ are singletons (hence compact) for $c > r$, but the entire $V$ for $c \le r$; the last one is compact iff it is finite iff either $V=0$ or ($F$ is finite and $\dim_F(V)$ is finite).)
Note now that the equivalence hold in great generality, but obviously neither of the statements is true in many settings. E.g. if $F$ itself is not locally compact, like $\mathbb Q$ with either the standard archimedean or any $p$-adic value, both statements are wrong e.g. if $\dim_F(V) \in \mathbb N_{\ge 1}$ (and for many infinite-dimensional $V$ as well).