I think the following meets your setup. Let $\mathbb K = \mathbb Z_2$ with the "absolute value" $|0|=0, |1|=1$ (this is non-Archimedean). See http://en.wikipedia.org/wiki/Absolute_value_%28algebra%29
Set $V = \mathbb Z_2^I$ for some index set $I$, with the trivial norm $\|0\|=0$ and $\|x\|=1$ for all other vectors $x$. This satisfies the usual rules, with $\|kx\| = |k|\|x\|$ for $k\in\mathbb K, x\in V$. Clearly $V$ actually have the discrete metric, and so is complete. Now, the closed unit ball is all of $V$, and not compact if $I$ is infinite.
BUT, I could instead define $\|x\|=2$ for $x\not=0$. Still we have a norm. Now the closed unit ball is $\{0\}$; and so is compact. All non-trivial linear maps have norm $1$.
To me, this seems like a very, very silly example, which perhaps shows that the original question needs tweaking with a bit...
If you start with an Archimedean absolute value, then really you have a subfield of $\mathbb C$ or $\mathbb R$ (which must contain $\mathbb Q$). By continuity, I then think you can turn $V$ into a $\mathbb R$ vector space, with a "norm" in the usual sense. Then if $V$ has a compact unit ball, it must be finite dimensional (over $\mathbb R$). So your example of $\mathbb R^2$ over $\mathbb Q$ is in a sense all that can happen.
You are misunderstanding Sternberg's argument. The argument goes like this (with my notations):
Consider two representations $V$ and $W$ of $G$ with the same character.
For any representation $S$ of $G$, let $\chi_S$ denote the character of $S$.
Take any decomposition $V=V_1\oplus V_2\oplus ...\oplus V_k$ of $V$ into irreducible subrepresentations.
Take any decomposition $W=W_1\oplus W_2\oplus ...\oplus W_l$ of $W$ into irreducible subrepresentations.
For any irreducible representation $P$ of $G$, we have $\left(\chi_P,\chi_V\right) = \left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$ and $\left(\chi_P,\chi_W\right) = \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$. Since $\chi_V = \chi_W$, we thus have
$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$
$= \left(\chi_P,\chi_V\right) = \left(\chi_P,\chi_W\right)$
$= \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$
for every irreducible representation $P$ of $G$. In other words, for every irreducible representation $P$ of $G$, the two lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same amount of entries isomorphic to $P$. In other words, the lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same entries up to isomorphism the same number of times (but of course, not necessarily in the same order). As a consequence of this, $V_1\oplus V_2\oplus ...\oplus V_k \cong W_1\oplus W_2\oplus ...\oplus W_l$. In other words, $V\cong W$, qed.
Nowhere did this use that the $V_i$ are somehow unique. (They are not unique as subspaces of $V$. They are unique up to isomorphism, and that follows either from the Krull-Remak-Schmidt theorem or from the fact that
$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right) = \left(\chi_P,\chi_V\right)$.)
Best Answer
For Banach spaces, the question is no : there is a form of the Peter-Weyl theorem, due to Shiga, which implies that in every Banach space representation of a compact group, the finite-dimensional sub-representations span a dense subspace. In particular, strongly continuous irreducible representations on a Banach space are finite-dimensional. I am not sure about arbitrary locally convex topological vector space.
Shiga's paper is here. If the link does not work, the precise reference is:
K. Shiga, Representations of a compact group on a Banach space, Journal of the Mathematical Society of Japan 7 (1955), 224–248.