Translations preserve compactly supported continuous functions

The maps $x\mapsto ax$, $x\mapsto x{a}^{-1}$ and $x\mapsto x^{-1}$ have continuous inverses $x\mapsto a^{-1}x$, $x\mapsto xa$ and inversion. They therefore carry the closure of the nonzero set onto the closure of its image. This gives exactly the three stated support formulas, and continuity of the pullbacks and compactness of their supports follow.

If $K=\mathrm{\varnothing }$, then $f=0$ and every asserted norm difference is zero. Otherwise choose a compact neighbourhood $V$ of $a_0$ and an open $O$ with $a_0\in O\subseteq V$. The sets $VK$ and $K{V}^{-1}$ are compact images of compact products. Their union $C$ is compact: restrict any open cover to each of the two sets and join the finite subcovers. For $a\in V$ it contains all left and right supports at $a$ and $a_0$.

Fix $ϵ>0$. The two jointly continuous functions $f(a^{-1}x)-f(a_0^{-1}x)$ and $f(xa)-f(x{a}_0)$ vanish at $(a_0,x)$. Collect all open rectangles about such points on which both absolute values are $<ϵ\mathrm{/}2$. Their second factors cover $C$, so finitely many suffice. Intersect their first factors with $O$, obtaining a neighbourhood $N$ of $a_0$. For $a\in N$ both differences are $<ϵ\mathrm{/}2$ throughout $C$ and vanish off $C$. Both uniform norms are therefore $\le ϵ\mathrm{/}2<ϵ$, as required. Only finite selections occurred.