Is there a direct proof that a compact unit ball implies automatic continuity?

One of the fundamental theorems in functional analysis is that if $X$ is a Banach space (say over $\Bbb C$) with a compact closed unit ball, then $X$ is finitely dimensional.

The usual proof is by assuming $X$ is infinite dimensional, and constructing by induction a sequence of vectors on the unit sphere which are not only linearly independent, but also have distances $>\frac12$ from one another.

But you can also prove this from automatic continuity. Namely, if every linear functional $f\colon X\to\Bbb C$ is continuous then $X$ has a finite dimension. If $\{v_n\mid n\in\Bbb N\}$ are linearly independent and lie the unit sphere, the function $f(v_n)=n$ can be extended to a linear functional on $X$. It is unbounded and therefore not continuous.

Can you prove directly from the assumption that the closed unit ball (equiv. the unit sphere) is compact that every linear functional is continuous?

This is definitely not an answer to the question. But it's a possibly amusing proof of the result - I was thinking about proving that automatic continuity and found I'd proved that $X$ is finite-dimensional.

Say $B(x,r)$ is the closed ball in $X$ about $x$ of radius $r$. Suppose $B(0,1)$ is compact. Then there exists a finite set $F\subset B(0,1)$ such that $$B(0,1)\subset\bigcup_{y\in F}B(y,1/2).$$

Hence, by the normed-vector-spaceness of $X$, for every $x\in F$ we have $B(x,1/2)\subset\bigcup_{y\in F}(x+B(y/2,1/4))$, so that $$B(0,1)\subset\bigcup_{y_0,y_1\in F}B(y_0+ y_1/2,1/4).$$Etc. It follows that for every $x\in B(0,1)$ there exist $y_0,y_1\dots\in F$ with $$x=\sum_{n=0}^\infty2^{-n}y_n.$$

And now if you regroup the terms in that sum it follows that $x$ is a linear combination of the elements of $F$, qed. (In fact $x/2$ is in the convex hull of $F$.)