So I'm just learning about the compact-open topology and am trying to show that for a compact, Hausdorff space ,$X$, the group of homeomorphisms of $X$, $H(X)$, is a topological group with the compact open topology. This topology has a subbasis of sets $\{f\in H(X):f(C)\subseteq V\}=S(C,V)$ for compact $C$, open $V$. This is my first attempt at getting to know this topology, so I'd appreciate some help with the part of a proof I have so far, possibly a better way to go about proving this, and any other help understanding this topology.
First, let $c:H(X)\times H(X)\rightarrow H(X)$ be composition. For $f,g\in H(X)$, let $S(C,V)$ be a subbasis set with $f\circ g\in S(C,V)$. So $f(g(C)\subseteq V$ which means $f\in S(g(C),V)$ and $g\in S(C,f^{-1}(V)$. The product of these open sets works since if $h_1(C)\subseteq f^{-1}(V)$ and $h_2(g(C))\subseteq V$, then $h_2\circ h_1(C)\subseteq V$.
Then let $i:H(X)\rightarrow H(X)$ be inversion. Take a subbasis set $O=\{g:g(C)\subseteq U\}$ and consider $i^{-1}(O)=\{g^{-1}:g(C)\subseteq U\}$. If $h^{-1}\in i^{-1}(O)$, then one thing we have is that $h(C)\subset U$, but I'm not totally sure what to do with this. This part seems like it should be easier, but I am just not seeing it.
Thanks.