Is there in general a connection between the Haar measure of a topological group and that one of its closed subgroup (with the induced topology)?
I know that for $O(n)$ with the induced standard topology and for the permutation matrices equipped with the discrete topology, this would not work. (One would get a zero measure, but the Haar measure must be nontrivial), but perhaps one can define the Haar measure of the subgroup by approximating sets by above or similar.
Let $G$ be a locally compact group, and let $\mu$ be a left Haar measure on $G$. A closed subgroup $X$ of $G$ has $\mu(X) > 0$ if and only if $X$ is also open. Then the restriction of $\mu$ to $X$ is a left Haar measure for $X$.
For a non-open (but still closed) subgroup $X$, we have $\mu(X) = 0$. Let $\nu$ be a Haar measure on $X$. Translates of $\nu$ will give you measures $\nu_Q$ on the cosets $Q \in G/X$. (Check: translations on which side??) Then (at least in the second-countable case) there should be a measure $\tau$ on $G/X$ so that we have a disintegration of $\mu$ of the form
$$ \mu(E) = \int_{G/X} \nu_Q(E)\;d\tau(Q) $$
(Without checking today I would conjecture that ) this is explained in:
Hewitt, E.; Ross, K. A., Abstract harmonic analysis. Volume I: Structure of topological groups, integration theory, group representations., Grundlehren der Mathematischen Wissenschaften. 115. Berlin: Springer- Verlag. viii, 519 p. (1994). ZBL0837.43002.