i think you mean banach space, not hausdorff space, and then this is a well known theorem of riesz: th.5.9.4., p.109, of dieudonne's foundations of modern analysis.
It is indeed true that a locally compact Hausdorff topological vector space E is finite dimensional.
Proof: Let K be a compact neighborhood of 0. We can assume that K is balanced. Since (1/2)K is a neighborhood of 0, there are finitely many points x_{1},...,x_{n} such that
Let M be the finite dimensional subspace spanned by the x_{1},...,x_{n}. Then M is closed. The quotient space E/M is Hausdorff. Since [itex]K\subseteq M+\frac{1}{2}K[/itex], then [itex]\varphi(K)\subseteq \frac{1}{2}\varphi(K)[/itex]. So (by induction) [itex]\varphi(2^nK)\subseteq \varphi(K)[/itex].
K is balanced, so [itex]E=\bigcup_n 2^n K[/itex]. Thus [itex]\varphi(E)=\varphi(K)[/itex]. Thus E/M is compact. Which implies that E/M is one point. Thus E=M.
what do you mean by zero in a topological space? oh i see micromass, you assumed he meant a vector space. the argument you gave is a variation of the argument for Riesz's theorem that I referred to. That's a nice example of extending an argument to a more general setting. I didn't know that version.