I'm trying to read this proof from Dow's An Introduction to Applications of Elementary Submodels to Topology. (Proposition 3.2.)
Let $M$ be a elementary submodel of large enough $H(\Theta)$ such that $M$ is $\omega$-covering, $(X,\tau)\in M$, and $|M|=\aleph_1$. I understand how proving that $\{U\cap M\mid U\in \tau\cap M\}$ generates the subspace topology for $X\cap M$ gives the result. So I just need help proving this.
The proof of this fact goes like this: Take $U$ an open set of $X\cap M$ with the subspace topology. There's a countable dense subset of $X\cap M\setminus U$ call it $D$, as $M$ is $\omega$-covering there's a countable $D'\in M$ such that $D\subset D'$. Here Dow says that it follows from $M\models D'\cup \{x\}$ is second countable, that there's a set $T\in\tau\cap M$ such that $x\in T$ and $T\cap D'\subset U$.
I haven't been able to verify this last implication.
Thank you for your help.
There's a countable $B\in M$ such that $M\models B$ is a base for $D'\cup \{x\}$. But, being countable, $B$ is actually a subset of $M$. Now, as $$H(\theta)\models B \text{ is a base for }D'\cup \{x\}.$$ and $U\cap D'$ is an open neighbourhood of $x$ in $D'\cup \{x\}$, there's a $T\in B$ such that $$x\in T\cap D' \subset U\cap D'\subset U.$$ But this $T$ is actually in $M\cap\tau$ because $B$ is a subset of $M$.
The point here is that $M$ can't talk about $U$ because it doesn't know it exists, but $H(\Theta)$ can. And as $B$ is exactly the same set in $M$ and $H(\Theta)$, the open set $H(\Theta)$ finds is also in $M$.