Well, the statement is true when $H$ is a Lie subgroup of $G$, not a simple subgroup. It is enough, to this end, that $H$ is a closed subgroup of $G$.

Concerning the proof, it is technical, but the general idea can be grasped from the following heuristic reasoning. $G$ acts transitively on $M:= G/H$ by representation $G \ni g \mapsto f_g$ where $f_g : M \to M$ is defined as:

$f_g : M \ni [g'] \mapsto [g'g]$

You can easily see that $f_g= id$ if and only if $g \in H$. Therefore, if you fix $p \in M$ you can reach every other point $q$ by means of a suitable $f_g$ using as degrees of freedom the $n = dim G$ coordinates determining $g$. This can be done completely disregarding the $m= dim H$ coordinates defining elements in $H$, since they give rise to trivial actions in $M$ which leave $p$ fixed. This means that, to define a point $q$ on $M$ (starting form a fixed on $p$) you need to fix $n-m$ coordinates. Thus $M$ has dimension $n-m = dim G - dim H$.