It's an important fact that locally compact Hausdorff topological vector spaces are finite-dimensional, a proof can be found here. I'm somewhat stuck with the proof.
If ${U}$ is a neighbourhood of the origin. then for every ${x \in V}$ we see that ${2^{-n} x \in U}$ for sufficiently large ${n}$. By compactness of ${K}$ (and continuity of the scalar multiplication map at zero), we conclude that ${2^{-n} K \subset U}$ for some sufficiently large ${n}$.
Here $K$ is a compact neighborhood of the origin, From the above argument, I know that for every $x\in K$, we have $x\in 2^n U$ sufficiently large $n$, then by the compactness of $K$, we can find a finite collection of sets $\{2^{n_1}U,\cdots,2^{n_k}U\}$ such that $$K\subset 2^{n_1}U\cup\cdots\cup 2^{n_k}U.$$ But how can we say there exists a $n$ such that $K\subset 2^nU$?
You can use the following fact - Every neighbourhood of $0$ contains a balanced neighbourhood of $0$. (Rudin Functional analysis theorem 1.14)
By a balanced set we mean a set $B$ such that $\alpha B\subseteq B$ for every scalar $\alpha$ such that $|\alpha|\le1$.
So WLOG let $U$ be balanced.
Now if $s\le t\Longrightarrow\dfrac{s}{t}\le1\Longrightarrow\dfrac{s}{t}U\subseteq U\Longrightarrow sU\subseteq tU$.
That is all that you need here.