This is a continuation of this question about continuity of inversion with the topology of uniform convergence.
For $M$ a bounded metric space, let $H_M$ be the space of homeomorphisms of $M$ with the supremum metric $$d_{H_M}(f, g) = \sup_{x \in K} d_M(f(x), g(x)).$$
Now let $S$ be a dense subset of a compact metric space $K$, and suppose we know that inversion $f \mapsto f^{-1}$ is a continuous function $H_K \to H_K$.
Question: Does it follow that inversion is a continuous function $H_S \to H_S$?
I keep getting stuck trying to prove this since it is not guaranteed that a homeomorphism of $S$ continuously extends to a homeomorphism of $K$.
Suppose that every homeomorphism $f$ of $S$ does continuously extend to a homeomorphism $\tilde f$ of $K$. If $(f_n) \to f$ is a convergent sequence in $H_S$, then $(f_n)$ converges uniformly to $f$ as functions $S \to S$, so then $(\tilde{f_n}) \to \tilde f$ uniformly as functions $K \to K$, so $(\tilde{f_n})$ converges to $\tilde f$ in $H_K$, so since inversion is continuous in $H_K$ we then have $(\tilde{f_n}^{-1})$ converging to ${\tilde f}^{-1}$ in $H_K$, so $(f_n^{-1})$ converges to $f^{-1}$ in $H_S$ by restriction to $S$. Of course this proof depends on the extensions of homeomorphisms from $S$ to $K$.