$\def\R{{\mathbb R}}
\def\N{{\mathbb N}}$
May I please receive help with the following problem? Thank you!
Assume our space is Hausdorff and non-empty. Set $S$ in $(X,T)$ is a $G_\delta$ set if its an intersection of countably many open sets of $X$.
(i) Let $(X,d)$ be a metric space. Show any closed set in $X$ must be a $G_\delta$ set in $X$.
$\textbf{Solution:}$ Let $F$ be any closed set in a metric space $(X,d)$. Let $\displaystyle{F_{\frac{1}{n}}:= \bigcup_{x\in F} B(x, \frac{1}{n})}$ where $B(x,\frac{1}{n})$ is open ball of radius $\frac{1}{n}$. Then, it is clear $F_{\frac{1}{n}}$ is open. Also, $F \subseteq F_{\frac{1}{n}}$ so $\displaystyle{F \subseteq \bigcap_{n=1}^\infty F_{\frac{1}{n}}}$. Again, let $\displaystyle{y \in \bigcap_{n=1}^\infty F_{\frac{1}{n}}}$. So, $y\in F_{\frac{1}{n}}$ for all $n\in \N, y\in B(x,\frac{1}{n})$ for some $x\in F$.
So $d(x,y) < \frac{1}{n} < \epsilon.$ Thus, $y$ is a limit point of $F$. Since, $F$ is closed, $y\in F.$ So $\displaystyle{F = \bigcap_{n=1}^\infty F_{\frac{1}{n}}}$ implies $F$ is $G_\delta$ set.
(ii) Let $X = \R^\R$ with product topology. Show any nonempty $G_\delta$ set in $X$ is uncountable. Specifically, if $p\in X$, then $\{p\}$ is not a $G_\delta$ set.
$\textbf{Solution:}$ Let $F$ be any non-empty $G_\delta$ sets in $\R^\R$ with product topology. Then, $\displaystyle{F := \bigcap_{n=1}^\infty F_n}$ where for all $n\in \N, F_n$ is open in $\R^\R$. We have to show that $F$ is uncountable. So, let $F$ be countable. Then, there exists some $m\in\N$ such that $F_n$ is countable. Moreover, $F_m$ is of the form $\displaystyle{F_m :=\prod_{j\in\R} U_j^{(m)}}$ where $U_j^{(m)}$ is open in $\R$. This implies $F_m$ contains uncountable elements, implying that we get a contradiction. Hence, $F$ is uncountable.
Best Answer
Your first proof is good.
Your second proof doesn't work. In particular, you seem to think that a countable intersection of uncountable sets must be uncountable. This is not true. For example, let $A_n = \{-1\} \cup [n, \infty)$, for $n \in \Bbb{N}$. Then $\{-1\} = \bigcap_{n \in \Bbb{N}} A_n$, despite each $A_n$ being uncountable.
You also seem to not be using the definition of the product topology correctly. Open sets in the product topology need not take the form $\prod_{r \in \Bbb{R}} U_r$. Instead, sets like these form a basis for the topology, i.e. we can express every open set $U$ as a union of sets like these.
However, we actually need another restriction! The basis is formed of sets of the form $\prod_{r \in \Bbb{R}} U_r$, where $U_r = \Bbb{R}$ for all but finitely many $r \in \Bbb{R}$. Without this restriction, we get the box topology, and singletons are indeed $G_\delta$ in the box topology. So, we'll need to make use of this fact.
Now, suppose $\{p\}$ is $G_\delta$. Then we can write $$\{p\} = \bigcap_{n \in \Bbb{N}} U^n,$$ where $U^n$ is open. Thus, we have $p \in U^n$ for all $n$. For each $n$, we can find a basic set in $U^n$ that contains $p$, i.e. we can find open sets $U_r^n \subseteq \Bbb{R}$, for all $r \in \Bbb{R}$, with equality for all but finitely many $r$, such that $$p \in \prod_{r \in \Bbb{R}} U_r^n \subseteq U^n.$$ Now, consider $$V_r = \bigcap_{n \in \Bbb{N}} U^n_r \subseteq \Bbb{R}.$$ Note that there's only countably many $r$ such that, for some $n$, $U^n_r \neq \Bbb{R}$. Thus, $V_r = \Bbb{R}$ for all but countably many $r \in \Bbb{R}$. Let $$V = \prod_{r \in \Bbb{R}} V_r.$$ Then $V \subseteq U^n$ for all $n$, hence $V \subseteq \{p\}$. However, $V$ is uncountable, since at least one (in fact, uncountably many) of its factors are $\Bbb{R}$. This is a contradiction, hence $\{p\}$ is not $G_\delta$.