Continuous ergodic measure preserving transformation thus transitive

Consider the probability space $$(X,\mathcal{B},\mu)$$ where $$X$$ is a compact metric space and $$\mu(A) > 0$$ for all $$A$$ non-empty. Let $$T:X \to X$$ be a continuous ergodic measure-preserving transformation.

By a theorem in my course, if $$A,B \in \mathcal{B}$$ are open and non-empty, then we have that $$\lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n - 1} \mu(T^{-k}A \cap B) = \mu(A)\mu(B) > 0$$ thus there exists $$m \in \mathbb{N}$$ such that $$\mu(T^{-m}A \cap B) > 0$$ and so $$T^{-m} A \cap B \neq \emptyset$$. Now I am trying to show that $$T$$ is transitive therefore need that there exists $$l \in \mathbb{N}$$ such that $$T^{l} A \cap B \neq \emptyset$$. This is close to what I have derived so far but I am stuck trying to understand how to get to the final result.

You have that for a $$m$$, $$\mu(T^{-m}A \cap B) > 0$$.
As the set $$T^{-m}A \cap B$$ has strictly positive measure then it is non empty that is $$T^{-m}A \cap B \ne \emptyset$$. Then taking $$T^m(T^{-m}A \cap B)=A \cap T^m(B)$$, you find that $$T^{m}B \cap A \ne \emptyset$$.
Changing the role of $$B$$ and $$A$$, you find what you want.