For 2. $K(X)$ is Polish, the intersection of countably many dense open sets is dense and hence nonempty, so there are plenty of $K$ on which $f$ is injective.
For 1. I would try something like this: take a countable base, $\mathcal{B}$, for the topology of $X$. For a pair $\langle B, C\rangle$ of elements of $\mathcal{B}$ with disjoint closures let $$U(B,C)=\{K \in K(X): f[K\cap B]\cap f[K\cap C]=\emptyset\}\text{.}$$
Then prove that $U(B,C)$ is open and dense.
First of all note that if $X$ is a Polish space (completely metrizable and separable space), and $F\subseteq X$ is closed, then $F$ is also Polish (usually with a different metric). The result you want can be shown to hold with a combination of the following two classical results:
Theorem 1: Let $X$ be a nonempty perfect Polish space (perfect means without isolated points), then there is an embedding of the cantor space $C$ into $X$.
Theorem 2 (Cantor-Bendixson): Let $X$ be a Polish space. Then $X$ can be written as a disjoint union $P\cup S$, where $S$ is countable open and $P$ is perfect.
Those results and much more material concerning Polish spaces can be found in the book by Kechris Classical Descriptive Set Theory, I'll write down the proofs for completeness, which I took from the aforementioned book.
To prove Theorem 1 we need the concept of a Cantor scheme on a space $X$. A family of subsets $(A_s)_{s\in 2^{<\Bbb N}}$ of $X$ is called a Cantor scheme if:
- $A_{s0}\cap A_{s1}=\varnothing$ for all $s\in 2^{<\Bbb N}$
- $A_{si}\subseteq A_s$ for all $s\in 2^{<\Bbb N}$ and all $i\in\{0,1\}$
Intuitively this is a way to split up the space that mimics the middle-third construction of the Cantor set.
Proof of Theorem 1: Let $X$ be a perfect Polish space. We will construct a Cantor scheme $(A_s)_{s\in 2^{<\Bbb N}}$ on $X$ satisfying:
- $A_s$ is open and nonempty for every $s\in 2^{<\Bbb N}$.
- $\mathrm{diam}(A_s)\leq 2^{-|s|}$.
- $\overline{A_{si}}\subseteq A_s$ for all $s\in 2^{\Bbb N}$ and all $i\in\{0,1\}$
Then, identifying $C$ with the product space $\{0,1\}^\Bbb N$, we define a map $f\colon C\to X$ by letting $f(c)=\bigcap_{n\in\Bbb N} \overline{A_{c|n}}$, which is a singleton because it is a decreasing intersection of closed sets with diameter going to $0$ in a complete metric space. Checking that $f$ is continuous and injective is easy, and any continuous injection from a compact space to an Hausdorff one is an embedding. We only need to construct such a Cantor scheme, which is easily done by induction on $|s|$. For $s=\varnothing$ let $A_\varnothing$ be an arbitrary nonempty open set in $X$. Given $A_s$ we can construct $A_{s1}$ and $A_{s0}$ by picking distinct elements of $A_s$ ($A_s$ cannot be a singleton since $X$ is perfect) and letting $A_{s1}$ and $A_{s0}$ be small enough balls around those two elements.
Proof of Theorem 2: Let $$P=\{x\in X\mid \text{every open nbhd of $x$ is uncountable}\}, $$ and let $S=X\setminus P$. $C$ is open countable because if $\{B_n\}_{n\in\Bbb N}$ is a countable basis of the topology of $X$, then $C$ is the union of those $B_n$ which are countable. To see that $P$ is perfect let $x\in P$. Then every nbhd of $x$ is uncountable, but then it must contain uncountably many points in $P$, since $X\setminus P$ is uncountable, so $x$ is not isolated in $P$ and $P$ is perfect.
Best Answer
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 $\mathcal{O}$.