A compact non-discrete group for any norm cannot be countable Let be $G$ a compact non-discrete group for some norm $N$, then $G$ is a Baire space.
Let us suppose that $G$ is countable.
Then:
$\begin{equation*}
G = \bigcup\limits_{x \in G} \{ x \}
\end{equation*}$
We have:


*

*$\{ x \}$ has no interior and is closed.

*$G$ is a countable reunion of closed sets with no interior.


By Baire theorem, $G$ has no interior.
But, as $G$ is closed and open by definition, $G$ must be its own interior.
We have a contradiction, then $G$ cannot be countable.
I am trying to show that a compact group for any norm cannot be countable, is this proof right? Can it be generalized further? Is there any useless hypothesis? Do I have the right to use the induced topology for $G$ where $G$ is something like $\mathrm{GL}_n(\mathbb{R})$?
 A: You're right that a compact topological group cannot be countably infinite, because of Baire (which implies that a countable Baire (Hausdorff top.) group is discrete, as there must be an open singleton, which implies that all singletons are open, etc.)
A better proof: 

Suppose $X$ is be a countable Baire topological group (which includes $T_1$-ness). Then $X$ is discrete.

Then $X = \cup_{x \in X} \{x\}$. The Baire theorem tells us that one $\{x\}$ (which is closed) must have non-empty interior, so such an $x$ is an isolated point. But a topological group is homogeneous, so this means that all $x \in X$ are isolated points. So $X$ is a discrete space.

Corollary: An infinite countable topological group cannot be compact.

In this context, compact implies Baire and an infinite discrete space is not compact.
The countably infinite is essential, as finite discrete groups are compact, and $S^1$ (any many others) is an uncountable compact group. In the first statement we cannot drop Baire, as the rationals are a non-discrete countable non-Baire group. And $\mathbb{Z}$ is the main (only?) example of a Baire countably  infinite topological group (locally compact implies Baire).
A: A compact topological group cannot be countable because it admits an Haar measure.
You have $1=\mu(G)=\sum_{g\in G} \mu(\{g\})$. It follows that the sum has to be finite (because $\mu$ gives the same measure on every singleton).
This is not true for non-compact groups (see my comment). For example $\mathbb{Z}$ with discrete topology is a topological Hausdorff group.
