I am trying to show the well-known fact that if a profinite group $G$ is not finite, then is not countable (i.e. $|G|\geq 2^{\aleph_0}$). There are many ways of proving this fact but I am interested in the following reasoning.
Since $G$ is a profinite group we have that it is an inverse limit of finite groups with the discrete topology $$G=\lim\limits_{\substack{\longleftarrow\\ i\in I}} G_i\subseteq \prod\limits_{i\in I} G_i.$$
We also have that $I$ is infinite because $G$ is not finite.
I will assume that the maps $\varphi_{i,j}: G_i\longrightarrow G_j$ are surjective.
It follows that $|G_j||\ker\varphi_{i,j}|=|G_i|$. And since, $|\ker\varphi_{i,j}|\geq 2$, it follows that for any $i\geq j$, $|G_i|\geq 2|G_j|$.
From this, I would like to deduce that $|G|\geq |2^{\aleph_0}|$.
Given that $|G|\geq |G_i|$ for any $i\in I$, I believe it follows that $$|G|\geq 2^{|I|}.$$ Because for any $j\in I$, $$|G|\geq \prod\limits_{\substack{i\in I\\ i\geq j}} |\ker \varphi_{i,j}||G_j|\geq \prod\limits_{\substack{i\in I\\ i\geq j}} 2.$$
$|G|\geq 2^{|I|}$ clearly implies that $|G|\geq 2^{\aleph_0}$.
But I don't know if my argument is sufficient or even correct.
Any help would be welcome.