Any two components of a topological group are homeomorphic 
Let $G$ be a topological group. Prove that any two components of $G$ are homeomorphic.

My idea: let $C_x$ and $C_y$ be two components for some $x,y\in G$. Since $G$ is a group we can find $g\in G$ such that $gx=y$. Note that the map $\mu:G\to G$ given by $h\mapsto gh$ is a homeomorphism. So it suffices to show that $\mu(C_x)=gC_x=C_y$. Now this is where I'm a little unsure... I'd like to do the following:
$$
\begin{align*}
gC_x&=g\bigcup_{\substack{x\in C\\C\text{ connected}}}C\\
&=\bigcup_{\substack{x\in C\\C\text{ connected}}}gC\\
&=\bigcup_{\substack{gx\in C\\C\text{ connected}}}C\tag{???}\\
&=\bigcup_{\substack{y\in C\\C\text{ connected}}}C\\
&=C_y.
\end{align*}
$$
My question is about the $(???)$ part above. Is that the correct way to do it? It seems like simple reindexing but for some reason it's tripping me up thinking about it. Is it because $gC$ is a connected set containing $gx$?

What you did is essentially correct.
It's also not really necessary: suppose $C_x$ and $C_y$ are the components of $x$, resp. $y$. We have a homeomorphism $h;X \to X$ so that $h(x)=y$ (we only need that $X$ is homogeneous; the full force of being a topological group is not needed, this is also what you really use, anyway). As $C_x$ is connected and contains $x$, $h[C_x]$ is connected ($h$ is continuous) and contains $h(x)=y$, so $h[C_x] \subseteq C_y$ (by maximality). OTOH, $h^{-1}[C_y]$ is connected (as $h^{-1}$ is also continuous) and contains $x$ so $h^{-1}[C_y] \subseteq C_x$ (likewise by maximality) or $C_y \subseteq h[C_x]$ and so $h[C_x]= C_y$ by two inclusions.
So $h\restriction_{C_x}: C_x \to C_y$ is also a homeomorphism and indeed $C_x \simeq C_y$.
I only use the simple fact

The component $C_x$ of $x \in X$ is the maximal (by inclusion) connected subset of $X$ that contains $x$.

which often makes reasoning about components somewhat easier IMO.
