As requested in the comments, here is a proof in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.
Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.
So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.
Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.
Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\cap T^{-k}E))=\varnothing$, a contradiction.$\square$
In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.
But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.