Assume that $T: X \to X$ is a homeomorphism of a compact metric space to itself, and let $\mu$ be a $T$-invariant Borel probability measure on $X$. I want to show that for almost every $x \in X$ the forward and backwards orbits of $x$ have the same closures, that is $\overline{\{T^n x : n \geq 1 \}} = \overline{\{T^{-n} x : n \geq 1\}}$ for almost every $x$ and that the closures contain $x$.
So, I first want to start with the last statement, $x \in \overline{\{T^n x : n \geq 1 \}}$.
So, $X$ is totally bounded. Take $\epsilon_n = \dfrac1{2^n}$ so and the take one of the cover sets $B_1(x, \epsilon_1)$, we know now by Poincaré recurrence that there exists a subset $F_1$ of $B_1$ such that for $x_1 \in F_1$, $T^{n^1_k} x_1$ always lands in $F_1$. By compactness, passing to a subsequence if necessary, $T^{n^1_k} x_1$ converges to $y_1 \in \overline F_1$. $\overline F_1$ is compact and closed, hence totally bounded, so again I can find a cover with $\epsilon_2$ as the ball radius and I can pick the first ball and $x_2\in F_2$. So then I have $T^{n^2_k} x_2\to y_2$, if I continue like this (needs verification) I think I can find $x \in \bigcap \overline F_n$ (Cantor nesting lemma) and then $y_n$ should converge to $x$. Since I only did all of this countably many times the union should give me a set of full measure. (For the sequence I take the diagonal sequence)
This seems to me overly complicated and a bit fishy so I would like to know if I'm on the right track, am I missing something clear?
This is a homework assignment, so I would appreciate it if I only get hints.