$\def\R{{\mathbb R}}
\def\Rhat{{\widehat{\R}}}
\def\n{{\mathbf n}}
\def\x{{\mathbf x}}$
May I please receive help with the following problem? Thank you.
I wish to prove $F\subseteq\R^n$ is closed if and only if every convergent sequence $\x^{(k)}\in\R^n$ such that for all $k$, $\x^{(k)}\in F$
has its limit in $F$, that is, if $\x^{(k)}\to\x$ then $\x\in F$.
$\textbf{Solution:}$
$(\leftarrow)$ Assume, every sequence $\x^{(k)}$ converges in $F$. We wish to prove $F$ is closed. Let $F$ be not closed then there exists $\x$ which a limit point of $F$. Since $\x$ is a limit point of $F$, so, there exists a sequence $\x^{(n)}$ of $F$ such that $\x^{(n)} \to \x$ and $\x\notin F$, which a contradiction to the supposition. Hence, $F$ is closed.
$(\rightarrow)$ Assume that $F$ is closed. We wish to prove every sequence $\x^{(n)}$ of $F$ converges to $\x$ in $F$. By definition, $d(\x^{(n)}, \x) < \epsilon$ for all $n\ge N$; for all $n\ge N$, $\x^{(n)} \in B(\x,\epsilon)$ [*].
Next, assume $\x\notin F$ then $\x \in F^c.$ Since $F$ is closed then $F^c$ is open. So $\x$ is an interior point of $F^c$. Thus, there exists $\epsilon > 0$ such that $B(\x, \epsilon) \subseteq F^c$ [**].
However, from [*] and [**], $\x^{(n)} \in B(\x, \epsilon) \subseteq F^c$ for all $n\ge N$, implying $\x^{(n)} \in F^c$ for all $n\ge N$, which is a contradiction as $\x^{(n)} \in F$ for all $n\ge N$. So our supposition is wrong, implying $\x\in F$. Thus, every convergent sequence of $F$ converges in $F$.
Best Answer
You can give a general proof for any metric space $(X,d)$ just using that a set is closed iff it's complement is open and that a set is open iff each point is an interior point:
Suppose $F$ is closed. Let $x_n \to x$ where all $x_n \in F$. We need to show that $x \in F$ too. So suppose that $x \notin F$. As $F^\complement$ is open, $x$ is an interior point of it, so there is some $r>0$ such that
$$B(x,r) \subseteq F^\complement\tag{1}$$
Now apply the definition of convergence to $(x_n)$ for this $r$ to find $N \in \Bbb N$ such that
$$\forall n \ge N: d(x_n, x) < r\tag{2}$$
But then $(2)$ implies that $x_N \in B(x,r)$ so by $(1)$, $x_N \notin F$, contradiction as all $x_n$ are in $F$ by assumption. So $x \notin F$ cannot hold and so $x \in F$.
Conversely: suppose that $F$ is closed under sequence limits. We will show that $F^\complement$ is open, so suppose for a contradiction that some $x \in F^\complement$ is not an interior point of $F^\complement$. This implies
$$x \notin F \text{ and } \forall r>0: B(x,r) \nsubseteq F^\complement$$
or otherwise put, using the definitions:
$$x \notin F \text{ and } \forall r>0: \exists y \in F: d(x,y) < r\tag{3}$$
Now to construct a sequence, apply $(3)$ for each $r=\frac{1}{n}$ where $n=1,2,3,\ldots$, and we find $x_n \in F$ such that $d(x_n, x) < \frac{1}{n}$ for each $n$.
Now it's easy to see (why?) that this implies that $x_n \to x$, but then this contradicts the property of $F$: all $x_n \in F$ but their limit $x$ is not in $F$. So $x$ is in fact an interior point of $F^\complement$ and $F^\complement$ is open and $F$ is closed. QED.