Let $K$ be a complete field w.r.t non-trivial absolute value $|-|$ and $V$ a normed vector space over $K$. Let $||-||=u(-)$ be sup norm w.r.t a fixed basis of $V$ over $K$ and $||-||'=v(-)$ be another norm. Note that $k\in K,||kx||=|k|\cdot||x||$ and we have this similarly for $||-||'$.
It is clear that w.r.t the fixed basis $\forall x\in V,v(x)\leq c u(x)$ which indicates continuity of $v$ against topology defined by $u$. Then Lang said "Hence $v$ has a minimum on the unit sphere w.r.t $u$(by local compactness)."
$\textbf{Q:}$ How did Lang use local compactness to deduce minimum here? Here is my reasoning. Since absolute value is non-trivial, I can define $O$ and $\exists |a|<1$. Now $a^nO\cong O$ by multiplication. Since $K$ is locally compact, I can shrink $O$ by multiplication of $a^n$ into the compact neighborhood of $0$. Since $O$ is closed by definition, for some $n$, $a^nO$ is compact which forces $O$ compact by homeomorphism. Now boundary(i.e. unit sphere) is closed in compact set and hence it is compact. (This is rather elaborate reasoning by assuming $K$ is topological field to start with though it is not hard to demonstrate this is compatible with $|-|$.)
Alternatively one can realize compactness by direct application of $a^n U$ with $U$ being the set of norm 1 elements. This is trivially closed and use continuity argument of $a^n$ to shrink to a closed subset of compact neighborhood of $0$. Hence it is compact and achieve the minimum.