Let $(X, d)$ be a metric space, and $(x_n)_{n \in \mathbb{N}}$ a sequence in $X$, with $x_n \rightarrow x$ w.r.t. $d$ for some $x \in X$. Show that the set $\{x_n : n \in \mathbb{N}\}\cup\{x\}$ is closed in $(X, d)$.
I figured I should use $\overline{F} = F \cup F'$ to show the above set is closed. Let $(x_n)_{n \in \mathbb{N}} = F$. Then, all elements of the sequence are in $F$, which is equal to $\{x_n : n \in \mathbb{N}\}$. Clearly, the sequence (or $F$) has only one accumulation point, $x$. So, $F' = \{x\}$. This shows $\{x_n : n \in \mathbb{N}\}\cup\{x\} = \overline{F}$, and thus is closed.
Is this sufficient in proving the problem? Or would I have to show that $x$ is the only accumulation point as well (figured I don't have to because we know limits are unique)? Any advice would be very helpful. Thank you.
It's not even obvious that $F$ has an accumulation point. Indeed, it can have none: for instance, if the sequence is constant.
What's true is that if an accumulation point exists, then it is $x$.
Indeed, suppose $y$ is an accumulation point. Then, for every $\varepsilon>0$, there are infinitely many $m$ such that $d(y,x_m)<\varepsilon/2$. Also, there exists $\bar{n}$ such that $d(x,x_n)<\varepsilon/2$, for every $n>\bar{n}$.
In particular, we can choose $m$ such that $m>\bar{n}$ and $d(y,x_m)<\varepsilon/2$. Hence $$ d(y,x)\le d(y,x_m)+d(x_m,x)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon $$ Since $\varepsilon$ is arbitrary, we conclude $y=x$.
If $x$ is not an accumulation point of $F$, then $x\in F$ (prove it).