This is an attempt at a relatively mild generalization of what others have said:
Let $K$ be a field and $|\cdot|: K \rightarrow \mathbb{R}$ be a nontrivial absolute value on $K$.
$\bullet$ If $K$ is complete for $|\cdot|$, then $E(K)$ has the structure of a $K$-analytic Lie group in the sense of Serre. In particular it is a $K$-analytic manifold so has at least continuum cardinality.
$\bullet$ When $|\cdot|$ comes from a rank one valuation $v$, I suspect that even if $K$ is merely Henselian for $v$, then $|\cdot|$, $E(K)$ cannot be finitely generated.
Here is a proof in the case that the valuation is discrete and the residue field $k$ is infinite: standard arguments involving the formal group still give a filtration
$E(K) \supset E^0(K) \supset E^1(K) \supset E^2(K) \supset \ldots$
such that (by Hensel's Lemma) for all $n \geq 1$, $E^n(K)/E^{n+1}(K) \cong (k,+)$. (Just last night I noticed that Cassels's Lectures on Elliptic Curves has a beautiful, elementary take on this. He works with the case $K = \mathbb{Q}_p$ but the argument holds much more generally.) If $k$ is infinite, then its additive group is not finitely generated and thus $E(K)$, having a subquotient which is not finitely generated, is itself not finitely generated.