Analytic sets have perfect set property (Kechris)

As title says. I’m trying to learn some descriptive set theory but I don’t quite see this.

I want to use the following:

Given $X, Y$ Polish spaces, $f:X\to Y$ continuous, if $f(X)$ is uncountable there is a subset $K\subseteq X$ homeomorphic to Cantor space on which $f$ is injective.

I can reduce to the case where $f(U)$ uncountable for $U$ open in $X$, and Kechris says to show $\{K\in K(X) : f \text{ injective on K}\}$ is a dense $G_\delta$ set, in the Vietoris topology on $K(X)$ the compact subsets of X.

1. How do I show this? I suspect “Lusin schemes” might be useful but I don’t really understand this technology. Other approaches are also welcome.

2. Why does this give the result? Being $G_\delta$, this set is then Polish (right?), but why does this yield an uncountable K (which I understand would be sufficient)

Thank you