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 $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, 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 $B_n$, see ($*$) below for more details), 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.


($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.