Let ($E$, $\Vert \cdot \Vert$ ) be a normed vector space and let $C$ be a convex and compact subset of $E$ such that $0 \in C$. I claim $C$ is contained in a finite dimensional subspace of $E$.
To see this, let $V = \text{span} (C)$. Take $x$ in the interior of $V \cap C$ relative to $V$ and a closed ball $B$ centered at $x$ so small that $B' := B \cap V \cap C$ is contained in $V \cap C$. Then $B'$ is a compact ball of $V$, hence $V$ is finite dimensional.
The issue here is: can I guarantee that the interior of $V \cap C$ is nonempty relative to $V$? I think it is geometrically clear, but I couldn't find a proof.
A counterexample to your claim is the Hilbert cube, the set of all vectors in $x\in \ell^2$ such that $0\le x_n\le 1/n$ for all $n$. It is compact and convex, but is not contained in any finite-dimensional subspace.
A subset of a Banach space is compact if and only if it is closed, bounded, and "flat", where the latter means: for every $\epsilon>0$ there is a finite-dimensional subspace $M$ such that the set is contained in the $\epsilon$-neighborhood of $M$. See my blog post Compact sets in Banach spaces.