How can a closed set in a finer topology be closed in a coarser topology? In Jan Reimann's Lecture Notes on Descriptive Set Theory, he proves the Perfect subset property for Borel sets (that is, in a Polish space, every uncountable Borel set has a perfect subset) in the following way.

The Theorem 2.4:

Every uncountable Polish space contains a homeomorphic embedding
of the Cantor space.

If $\mathcal{O'}$ is finer than $\mathcal{O}$, then it should have more open and therefore closed sets. How can $\mathcal{O'} \supset \mathcal{O}$ result in every closed set in $(B,\mathcal{O'}|_B)$ is also closed in $\mathcal{O}$? Do I miss something here?
If this method to prove the closedness is actually wrong, is there a way to remedy the proof?
 A: I don't exactly follow this reasoning. Clearly the statement that every closed subset in $\mathcal{O'}$ is closed in $\mathcal{O}$ is wrong (otherwise these topologies would be simply equal). But I'm not even sure why the author looks at closed subsets at all, at this point.
It looks like the author got distracted somewhere, because the overall argument is fine. It just needs a bit of tweaking:
Since $\mathcal{O'}\supseteq \mathcal{O}$ then the continuity of $f:2^{\mathbb{N}}\to (B,\mathcal{O'})$ implies continuity of $f:2^{\mathbb{N}}\to (B,\mathcal{O})$. In particular $f(2^{\mathbb{N}})$ is closed (even compact) in $(B,\mathcal{O})$ and it has no isolated points as a homeomorphic image of the Cantor space. Therefore $f(2^{\mathbb{N}})$ is perfect with respect to $\mathcal{O}$.
A: Let us Clconsider $(\Bbb{R}, \tau_{\text{std}}) $ and the Borel set $[0, 1)$ .
Now consider the Lower limit topology $\tau_{\text{lower}}$ on $\Bbb{R}$
Then $[0, 1) $ is a clopen set in $(\Bbb{R}, \tau_{\text{lower}}) $.
Now consider $[0, \frac{1}{2}) \subset [0, 1) $ . Then $[0, \frac{1}{2})$ is closed in $([0, 1), \tau_{\text{lower}|_{[0,1)}}) $.
But $[0, \frac{1}{2})$ is not closed in $([0, 1), \tau_{\text{std}}|_{[0,1)}) $
