property $P$ type $a$ and irreducibility with respect to $P$ 
Let $(Y, \tau)$ a 2-countable topological space. It is said that a property $P$ is type $a$ if for each decreasing sequence of closed $\{A_n : n \in \mathbb{N}\}$ such that $A_n$ has the property $P$ for each $n \in \mathbb{N}$, the intersection $\cap_{n \in \mathbb{N}} A_{n}$ it also has the property $P$. We will say that $A \subseteq Y$ is irreducible with respect to $P$, if no proper and closed subset of $A$ has the property $P$. Show that if $P$ is type $a$ and some closed of $Y$ has the property $P$, then there exists a closed subset of $Y$ with the property $P$ that is irreducible with respect to the property $P$.

Suppose by contradiction. For everything closed $A \subseteq Y$ with property $ P $ there is a subset of its own and closed  $B_1 \subset A $ that has property $ P $, Furthermore, since $ B_1 $ is also closed of $ Y $ and has the property $ P $, there is a subset $ B_2 \subset B_1 $ closed of $ B_1 $ with the property $ P $. Inductively we can create a succession of decreasing closures. Is it possible to construct a sequence as we did previously such that the $\cap_{n \in \mathbb{N}} B_n$ does not have the property $ P $? is there another way to test it?

Suppose that there is no such irreducible set.
Let $A_0\subseteq Y$ be a closed set with property $P$. Suppose that $\alpha<\omega_1$, and for $\xi<\alpha$ we have constructed closed sets $A_\xi$ with property $P$ such that $A_\xi\subsetneqq A_\eta$ whenever $\eta<\xi<\alpha$. If $\alpha=\beta+1$, by hypothesis there is a closed $A_\alpha\subsetneqq A_\beta$ with property $P$. If $\alpha$ is a limit ordinal, there is a strictly increasing sequence $\langle\xi_n:n\in\omega\rangle$ of ordinals less than $\alpha$ such that $\alpha=\sup_{n\in\omega}\xi_n$, and we set
$$A_\alpha=\bigcap_{n\in\omega}A_{\xi_n}=\bigcap_{\xi<\alpha}A_\xi\,;$$
clearly $A_\alpha$ is closed, and it has property $P$ because $P$ is type $a$. Thus, the recursion goes through to $\omega_1$ to give us sets $A_\xi$ for $\xi<\omega_1$ such that each is closed and has property $P$, and $A_\xi\subsetneqq A_\eta$ whenever $\eta<\xi<\omega_1$.
Let $\mathscr{B}$ be a base for $Y$. For each $\xi<\omega_1$ there is a point $y_\xi\in A_\xi\setminus A_{\xi+1}$, and there is a $B_\xi\in\mathscr{B}$ such that $y_\xi\in B_\xi\subseteq Y\setminus A_{\xi+1}$. It follows that if $\eta<\xi<\omega_1$, then $y_\xi\in B_\xi\setminus B_\eta$, so $B_\xi\ne B_\eta$ and hence that $\mathscr{B}$ is uncountable, contradicting the second countability of $Y$.
