I think that every compact $\Pi^0_1$-class has a hyperarithmetic infinite path. 


>  **Proof**: Given a compact $\Pi^0_1$-class $P$. So there is a recursive tree
> $T\subseteq \omega^{<\omega}$ so that $[T]=P$. Since $P$ is compact,
> there is a function $f$ dominating all of the members of $P$. We will
> find a hyperaritmetic one.
> 
> We use Spector-Gandy's theorem to recursively work on
> $L_{\omega_1^{CK}}$.
> 
> For $n=0$, there is some number $m$ and stage $\alpha<\omega_1^{CK}$
> so that for every $k\geq m$, $T_k=\{k\sigma\mid k\sigma\in T\}$ is
> well founded witnessed at stage $\alpha$. Find the first such stage
> $\alpha_0$ and corresponded number $m_0$. Let $f(0)=m_0$.
> 
> Generally for any $n+1$, there is some number $m$ and stage
> $\alpha<\omega_1^{CK}$ so that for every $k\geq m$, $T_{f(0)\cdots
> f(n)k}=\{k_0\cdots k_n k\sigma \mid k_0\cdots k_n k\sigma\in T  \wedge
 \forall i\leq n (k_i\leq f(i))\}$ is well founded witnessed at stage
> $\alpha$. Find the first such stage $\alpha_{n+1}$ and corresponded
> number $m_{n+1}$. Let $f(n+1)=m_{n+1}$.
> 
> So $f$ is a total $\Pi^1_1$ function and so hyperarithmetic. Thus $P$ must contains a hyperarithmetic member since every member of $P$ is dominated by $f$. QED