= Solution
The canonical image of $\Gamma$ is <dense> in its <profinite completion> $\widehat\Gamma$. Therefore every $x\in\widehat\Gamma$ is a limit of a net $(\delta_\lambda)$ in $\Gamma$. Continuity of conjugation gives
$$
\delta_\lambda\gamma\delta_\lambda^{-1}
\longrightarrow x\gamma x^{-1},
$$
so every element of $\operatorname{Cl}_{\widehat\Gamma}(\gamma)$ lies in the <closure> of $\operatorname{Cl}_\Gamma(\gamma)$. The reverse inclusion follows because the larger conjugacy class contains the smaller one and is closed by part b(iii). Hence
$$
\boxed{\overline{\operatorname{Cl}_\Gamma(\gamma)}
=\operatorname{Cl}_{\widehat\Gamma}(\gamma)}.
$$
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