The Bernstein construction of a set without the perfect set property gives, from a $\Sigma^1_2$ well-order of the reals, a $\Sigma^1_2$ set without the perfect set property. What I'd like to know is whether this can be taken to be the well-ordering itself; for instance, does $<_L$ have a perfect subset (working in $L$)? Note that well-orders can have perfect subsets; see, for instance, the answer to this post.
More generally, can a well-order of $\mathbb{R}$ of minimal ordertype have a perfect subset? Note that the sort of counterexamples given in the previous answer will never have ordertype $2^{\aleph_0}$.
The answer might depend on how things are coded. Under a reasonable coding, it does contain a perfect subset (of the space ${^\omega}2\times{^\omega}2$). Suppose that for reals $x,y$ of the same constructibility rank, if $x(0)<y(0)$ then $x<_Ly$ (this is consistent with the usual definitions; the details of the Gödel coding etc are not usually spelled out in detail, and they could be arranged to do this). Then $<_L\upharpoonright{^{\omega}2}$ contains a perfect subset, because it contains all elements of the form $((0,1),(n_1,n_1),(n_2,n_2),(n_3,n_3),\ldots)$ where $\left<n_k\right>_{1\leq k<\omega}$ is an arbitrary element of ${^\omega}2$. (This is because this is a pair of reals $(x,y)$ where $x(0)<y(0)$ but $x(n)=y(n)$ for $n>0$, so $x,y$ have the same constructibility rank.) The set of all such elements of ${^\omega}2\times{^\omega}2$ is perfect.