I believe this fails under $V=L$. If $P$ is a pointed perfect set and $X$ is a real, let $P(X)$ be the element of $P$ where at every split we choose according to the next bit of $X$. So $X\oplus P \equiv_T P(X)$.
We'll build the $A_n$ via an $\omega_1$ length construction as follows. At stage $\alpha$, we consider $P_\alpha$ the next pointed perfect set, and we choose $X_0, X_1, \dots$ which are self-constructible reals of $L$-rank at least $\alpha$ and computing $P_\alpha$. We omit $P_\alpha(X_n)$ from $A_n$.
We define $A_n$ to consist of those reals which are never omitted. So $Y \not \in A_n$ iff there is a well-founded $L_\alpha \in \Delta^1_1(Y)$ realizing $Y$'s omission, which is $\Pi^1_1$. By appropriate choice of $X_n$, no real is omitted from more than one $A_n$, and so $\bigcup_n A_n = 2^\omega$. But clearly we have diagonalized against any $A_n$ containing a perfect pointed set.
