Perfect Kernel analogue for NON $\sigma-$compact subsets of a Polish space

I'm looking at the proof of the following theorem by Hurewicz (7.10 in Kechris' Descriptive Set Theory):

Suppose $$X$$ is a Polish Space. Then one of the following holds:

1. $$X$$ is $$K_\sigma$$ (i.e, $$\sigma$$-compact, i.e, a countable union of compact sets.

2. $$X$$ has a closed subset homeomorphic to $$\omega ^{\omega}$$

My question is about a very small step in the proof of this. We build a Lusin scheme as expected,and something that comes up in that process is the following:

If $$F \subseteq \omega ^ {\omega}$$ is not $$K_\sigma$$, then $$H:= \{x \in F: \forall U$$ open nbhd of $$x$$, $$\overline{U \cap F}$$ is not $$K_\sigma\}$$ (this is what I called the perfect kernel analogue)

Now the text quickly says "$$H$$ is nonempty since $$F$$ is not $$K_\sigma$$, and similarly $$F\setminus H$$ is contained in a $$K_\sigma$$ set", but I don't see why this is true. I must be missing something obvious given that the book skips over it so I apologize if it's super dumb.

My guess is: suppose $$H \not = \emptyset$$. Then since $$X$$ is separable, there is a countable set of points $$\{x_i\}$$ dense in $$F$$ with open neighborhoods $$U_i$$ such that $$\overline{U_i \cap F}$$ is $$K_\sigma$$. These $$U_i$$ cover $$F$$ and thus $$F$$ is a countable union of $$K_\sigma$$ sets and thus $$K_\sigma$$, yielding a contradiction. A similar idea can be used to show $$F\setminus H$$ is $$K_\sigma$$. I don't think this is entirely right though becuase the $$U_i$$ don't have to cover $$F$$ necessarily.

What is the right approach here? I feel like I'm missing something super obvious.

Your idea seems correct to me, except for a typo at the beginning where it should be suppose $$H=\varnothing$$ rather than $$H\neq\varnothing$$ to reach a contradiction.
You say that the $$U_i$$ don't have to cover $$F$$, but in fact they do since the $$\{x_i\}$$ are a dense subset.
Similarly pick a countable dense subset $$\{x_i\}$$ in $$F\setminus H$$. By definition of $$H$$ each of those points has a neighbouhood $$U_i$$ such that $$\overline{F\cap U_i}$$ is $$K_\sigma$$, thus $$F\setminus H\subseteq\bigcup_{i<\omega}\overline{F\cap U_i}$$ is contained in a $$K_\sigma$$ set, which means that $$H$$ cannot be compact, otherwise the whole of $$F$$ would be $$K_\sigma$$.