Compact implies Totally Bounded in a TVS Definition: If $X$ is a TVS and $E\subseteq X$ then $E$ is totally bounded iff for every nbhd of $0_X$, $V$, there exists some finite set $F\subseteq X$ such that $E\subseteq F+V$.
Claim: If $K$ is compact then $K$ is totally bounded.
Proof: Let $V$ be a given nbhd of $0_X$. Then $\{x+V\}_{x\in X}$ is an open cover of $X$, and hence of $K$. So $\exists$ some $\{x_i\}_{i=1}^{n}\subseteq X$ such that $K\subseteq \bigcup_{i=1}^n (x_i+V)$. Then $K$ is totally bounded.
Question: Is this proof correct? If yes, why does Rudin seem to imply that it is necessary for $X$ to be a Frechet space to make that claim (FA, pp. 73 (c))? 

Your proof is fine.
In addition, looking at the proof of Th. 3.20 (c) of Papa Rudin's Functional Analysis, he just mentioned for metric spaces, closure of totally bounded sets are compact. Applying this to a compact set in a Frechet set, you conclude that the closed convex hull of a compact set is again compact. For the conclusion he uses the property (b) ensuring convex hulls of totally bounded sets are again totally bounded - even in non-metrizable spaces.
So in fact, your argument varifies the "obviously" in the mentioned proof.
