Show any closed set in $X$ must be a $G_\delta$ set in $X$ as well as if $p\in X$, then $\{p\}$ is not a $G_\delta$ set. $\def\R{{\mathbb R}}
\def\N{{\mathbb N}}$ 
Can I please receive feedback on my solution for the following problem? Thank you.
Assume any space is Hausdorff and non-empty, and any examples are also Hausdorff and non-empty.
A set $S$ in $(X,T)$ is called a $G_\delta$ set if it is an intersection of countably many open sets of $X$. 
(i) Let $(X,d)$ be a metric space. Show that any closed set in $X$ must be a $G_\delta$ set in $X$. 
$\textbf{Solution:}$ Let $(X,d)$ be a metric space, and let $C\in X$ be a closed set. Then $$\displaystyle{C=\bigcap_{i=1}^\infty U_i \text{ where } U_i = \bigcup_{x\in C} B(x,\frac{1}{i})}$$ $$C= \bigcap_{i=1}^\infty [\bigcup_{x\in C}B(x,\frac{1}{i})].$$ Each $\displaystyle{\bigcup_{x\in C} B(x,\frac{1}{i})}$ is an open set and $\displaystyle{\{x\} = \bigcap_{i=1}^\infty B(x,\frac{1}{\delta})}$ is an open set. Hence, the result.
(ii) Let $X = \R^\R$ with the product topology. Show that any nonempty $G_\delta$ set in $X$ is uncountable. Particularly, if $p\in X$, then $\{p\}$ is not a $G_\delta$ set. 
$\textbf{Solution:}$ Product topology implies we have finitely many proper open sets, i.e., $$U_1 \times U_2 \times \dots \times U_n \times \R^{\R - n}.$$ Now, we have countable intersection of sets of these type. Hence, only countably many places are proper and we still have a $\R^\R$ in the tail. Hence, any non-empty $G_\delta$ set in $X$ is uncountable.
 A: In (i) you’ve not actually proved that $C=\bigcap_{i=1}^\infty U_i$; for this you really do need to use the fact that $C$ is closed. To see this, suppose that the space is $\Bbb R$ with the usual metric, and let $C=[0,1)$. Define the sets $U_i$ for $i\in\Bbb Z^+$ exactly as you did in your argument. For $i\in\Bbb Z^+$ let $x_i=1-\frac1{2i}$; then $1\in B\left(i,\frac1i\right)$, so $1\in U_i$. If your argument were correct, it would follow that $1\in C$, but in fact it is not.
To fix the argument, show that $\bigcap_{i=1}^\infty U_i$ is closed. Suppose that you’ve done this; then $\bigcap_{i=1}^\infty U_i$ is closed and certainly contains $C$, so it contains $\operatorname{cl}C$, and all that remains is to show that $\bigcap_{i=1}^\infty U_i\subseteq\operatorname{cl}C$, which is not too hard. To show that $\bigcap_{i=1}^\infty U_i$ is closed, show that $\operatorname{cl}U_{i+1}\subseteq U_i$ for each $i\in\Bbb Z^+$.
Your argument for (ii) is also seriously incomplete, because you’ve considered only open sets in the usual base for the product topology. An open set in $\Bbb R^{\Bbb R}$ can be the union of uncountably many of those basic open sets, and intersections of open sets can be very complicated indeed. One way to proceed is as follows. 
Let $C$ be a countable subset of $\Bbb R^{\Bbb R}$, and suppose that $C$ is a $G_\delta$ set; then there is a countable family $\mathscr{U}$ of open sets in $\Bbb R^{\Bbb R}$ such that $\bigcap\mathscr{U}=C$. Fix a point $p\in C$ and show that there is another countable family $\mathscr{V}$ of open sets such that $C\cap\bigcap\mathscr{V}=\{p\}$. Conclude that $\mathscr{U}\cup\mathscr{V}$ is a countable family of open sets whose intersection is $\{p\}$, so that $\{p\}$ is a $G_\delta$ set. Show that there is a countable family of basic open sets — sets that restrict on only finitely many factors — whose intersection is $\{p\}$, and argue as you did in your question to show that this is impossible. This contradiction shows that $C$ could not have been a $G_\delta$ set after all.
