For any non-empty bounded set $A \subseteq \Bbb R$, we define the diameter of $A$, $\operatorname{diam} A$, as $\sup A - \inf A$. Let $\mathfrak{P}$ be the set of all non-empty perfect subsets of $\Bbb R$.
There exists a function $f:\mathfrak{P} \times \Bbb N^+ \to \mathfrak{P}$ such that $f(P,n) \subseteq P$ and $\operatorname{diam}f(P,n) \le \dfrac{1}{n}$ for all $(P,n) \in \mathfrak{P} \times \Bbb N^+$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
For $(P,n) \in \mathfrak{P} \times \Bbb N^+$, there exists $m \in \Bbb Z$ such that $P \cap \left ( \dfrac{m}{n}, \dfrac{m+1}{n} \right ) \neq \emptyset$ (Otherwise, $P= \left \{\dfrac{m}{n} \mid m \in \Bbb Z \right \}$, so all points of $P$ would be isolated).
Take the least $m \ge 0$ with this property, or, if none exists, the greatest $m < 0$ with this property.
Let $\alpha = \inf P \cap \left ( \dfrac{m}{n}, \dfrac{m+1}{n} \right )$ and $\beta = \sup P \cap \left ( \dfrac{m}{n}, \dfrac{m+1}{n} \right )$.
Clearly, $\dfrac{m}{n} \le \alpha \le \beta \le \dfrac{m+1}{n}$. So $\beta-\alpha \le \dfrac{1}{n}$. We then define function $f:\mathfrak{P} \times \Bbb N^+ \to \mathfrak{P}$ by $$f(P,n) = P \cap [\alpha,\beta]$$
Clearly, $f(P,n)$ is closed and non-empty. The definitions of infimum and supremum guarantee that neither $\alpha$ nor $\beta$ is isolated point of $f(P,n)$.
Next we prove that for all $a \in P \cap (\alpha,\beta)$, $a$ is not an isolated point of $f(P,n)$.
Assume the contrary that there exists $a \in P \cap (\alpha,\beta)$ such that $a$ is an isolated point of $f(P,n) =$ $P \cap [\alpha,\beta]$. Then there exists $\delta >0$ such that $x \neq a$ and $|x-a| < \delta$ implies $x \notin P \cap [\alpha,\beta]$. Let $\delta' = \min \{\delta, \beta -a, a-\alpha\}$. It follows that $x \neq a$ and $|x-a| < \delta'$ implies $x \notin P \cap [\alpha,\beta]$. But for such $x$, $|x-a| < \delta' \implies \alpha < x < \beta \implies x \in [\alpha,\beta]$, so $x \notin P$, showing that $a \in P$ is also an isolated point of $P$. This contradicts the fact that $P$ is perfect.
I don't see logical flaws or gaps. Just small things: at the beginning there should be $P ⊆ \{\frac{m}{n}: m ∈ ℤ\}$ instead of equality, and maybe add brackets: $α = \inf\bigl(P ∩ (\frac mn, \frac {m + 1}n)\bigr)$.
Also note that you may define $f$ directly as $f(P, n) := \overline{P ∩ (p - \frac 1{2n}, p + \frac 1{2n})}$ where $p$ is any point of $P$ (e.g. the greatest point closest to zero), and you additionally obtain $f(P, n) ⊆ f(P, m)$ if $n ≥ m$.
Also in general, for every set $A$ without isolated points and an open set $U$, you have that $A ∩ U$ has no isolated points, so $P := \overline{A ∩ U}$ is perfect. Moreover, $\operatorname{diam}(P) ≤ \operatorname{diam}(U)$, and if $A$ was perfect, then also $P ⊆ A$. This works in any metric space.