[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to believe is the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong *in general* (see the previous comment by Liviu): you may need more time.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take  $U$ to be the set of odd/even numbers, then your probability is 0, since you will leave the set immediately after entering it.