(a) A $G_\delta$ set in a space $X$ is a set $A$ that equals a countable intersection of open sets of $X$. Show that in a first-countable $T_1$ space, every one-point set is a $G_\delta$ set.
My attempt:
Approach(1): Let $x\in X$. Since $X$ is first countable at $x$, $\exists \{ U_n \in \mathcal{N}_x|n \in \Bbb{N}\}$ with the following property: $\forall U\in \mathcal{N}_x$, $\exists m\in \Bbb{N}$ such that $U_m \subseteq U$. Claim: $\bigcap_{n\in \Bbb{N}} U_n=\{x\}$. Proof: $\{x\} \subseteq \bigcap_{n\in \Bbb{N}} U_n$ holds trivially. Conversely, assume towards contradiction, $\exists y\in \bigcap_{n\in \Bbb{N}} U_n \subseteq X$ such that $y\neq x$. Since $X$ is $T_1$ space, $\exists P,Q\in \mathcal{T}_X$ such that $x\in P$, $y\notin P$ and $y\in Q$, $x\notin Q$. So $P\in \mathcal{N}_x$, $\exists m\in \Bbb{N}$ such that $U_m \subseteq P$. $y\in U_m \subseteq P$. So $y\in P$. Which contradicts our initial assumption of $X$ is $T_1$ space. Thus $\bigcap_{n\in \Bbb{N}} U_n \subseteq \{x\}$. Hence $\bigcap_{n\in \Bbb{N}} U_n = \{x\}$. Is this proof correct?
Approach(2): Let $x\in X$. Since $X$ is first countable at $x$, $\exists \{ U_n \in \mathcal{N}_x|n \in \Bbb{N}\}$ with the following property: $\forall U\in \mathcal{N}_x$, $\exists m\in \Bbb{N}$ such that $U_m \subseteq U$. Since $x\in U_n$ $\forall n\in \Bbb {N}$, $\bigcap_{n\in \Bbb{N}} U_n \neq \emptyset$. It suffices to show $\bigcap_{n\in \Bbb{N}} U_n \subseteq \{x\}$. Let $y\in X$ with $y\neq x$. Since $X$ is $T_1$, $\{y\}$ is closed. $X-\{ y\}\in \mathcal{T}_X$ and $x\in X-\{ y\}$. So $X-\{ y\} \in \mathcal{N}_x$. $\exists m\in \Bbb{N}$ such that $U_m \subseteq X-\{ y\}$. Which implies $y\notin U_m$. Thus $y\notin \bigcap_{n\in \Bbb{N}} U_n$, $\forall y\in X-\{x\}$. $(\bigcap_{n\in \Bbb{N}} U_n) \cap (X-\{x\})= \emptyset$. Hence $\bigcap_{n\in \Bbb{N}} U_n \subseteq \{x\}$. Our desired result. Is this proof correct?
Slight Variation of Approach(2): suppose $y\in \bigcap_{n\in \Bbb{N}} U_n$ such that $y\neq x$. So $X-\{y\} \in \mathcal{N}_x$, $\exists m\in \Bbb{N}$ such that $U_m \subseteq X-\{y\}$. So $y\in U_m \subseteq X-\{y\}$. $y\in X-\{y\}$. Thus we reach contradiction.
Question: Is $G_{\delta}$ a set(collection of objects)? Like $G_\delta =\{ \bigcap_{i\in \Bbb{N}} A_i| A_i \in \mathcal{T}_X, \forall i\in \Bbb{N}\}$.