My intuition is that the answer is yes:
Let $G$ be the original group, and let $H$ be a subgroup of $G$.
Let $\mu$ be a Haar measure on $G$ that is both right- and left-invariant.
I think that if we restrict $\mu$ to $H$ and restrict the translation
to translations by elements of $H$, then invariance must be preserved. 
Thus, by hand-waving, I guess that $\mu$ must be both right- and left-invariant on $H$, and not just on $G$. 
The reason that I'm not sure of it is the following:
The 2x2 affine matrix group is not unimodular. It is, however, a subgroup of $GL(2)$.


Now, this may be just fine: Both the affine group and GL(2) are not connected. And, while $GL(2)\_+$ is unimodular, I think that $GL(2)$ is not unimodular. Thus, the fact that
the 2x2 affine group is not unimodular does not cause any problem. 
Can someone please verify I'm right on this?


If I'm right, this brings me to another question:
Is every connected matrix group unimodular? I think that, pending on a positive answer
to my original question, this must be the case since they are all subgroups of 
$GL(n)\_+$ for some $n$.


Any help is much appreciated. 


Thanks!



Is every subgroup of a connected unimodular (matrix) Lie group also unimodular?
Take the so-called ax+b group, i.e. the connected component of the affine group of the real line. Or, even more concretely,
$$\left\{ \left( \matrix{ a & b \\ 0 & 1 } \right) \colon a>0, b\in{\mathbb R} \right\}.$$
This is not unimodular.


So I think the "proof by handwaving'' has a mistake somewhere. Probably your conception of restricting a measure from a space to a closed subset may need rethinking ...


