A subgroup of full measure is dense given a haar measure I want to know why if $\mu$ is a haar measure on a compact $G$ and $\mu(A)=\mu(G)$ then $A$ is dense in $G$. This fact is mentioned in the wikipedia page, but I couldn't find a proof for it.

False as stated.   Let $G$ be $\mathbb Z^2$ with the discrete topology.  Haar measure is counting measure.  Let $A = \left\{(x,0) \mid x \in \mathbb Z\right\}$.  Then $A$ is closed, $\mu(A) = +\infty = \mu(G)$.  But $A$ is not dense in $G$.

To get the correct statement: replace $\mu(A) = \mu(G)$ by $\mu(G \setminus A) = 0$.
