Let $(\textbf{a}_{n})_{n = 1}^{\infty}$ be a sequence in $\mathbb{R}^k$. Show $\{\textbf{a}_{n} : n \geq 1 \} \cup \{\textbf{a}\}$ is closed. Let $(\textbf{a}_{n})_{n = 1}^{\infty}$ be a sequence in $\mathbb{R}^k$ and $$\lim_{n \rightarrow \infty}\textbf{a}_{n} = \textbf{a}.$$ Show $B = \{\textbf{a}_{n} : n \geq 1 \} \cup \{\textbf{a}\}$ is closed.
Edit: I would specifically like to use the limit point definition of closed.  i.e:


A set $A$ is considered closed if it contains all of its limit points


EDIT: I think I've gotten a way to prove it.
Attempt 2:
We will show $B^{c}$ is open. This means I have to construct an open ball $B_{\delta}(x)$ s.t $B_{\delta}(x) \subset B^{c}$.
Given that the sequence $a_{n}$ converges then  for all $\epsilon > 0$ there exists a $N_{\epsilon} > 0$ such that for all $n > N_{\epsilon}$, $\|a_{n} - a \| < \epsilon$.
Let $x \in B^{c}$, $\epsilon > 0$, and $\epsilon = \frac{|x-a|}{2}$.
As $a_{n} \rightarrow a$ then $\|a_{n} - a\| < \frac{|x-a|}{2}$, let $\omega = min \{\frac{|x-a|}{2}, \frac{|x-a_{1}|}{2}, \dots, \frac{|x-a_{n}|}{2}\}$.
This then means $\|x-a_{n}\| > \omega \\ \Rightarrow \ B_{\omega}(x) \subset B^{c}$
Therefore $B^{c}$ is open and this means $B$ is closed.
Comment:
I'm not sure if the ball I constructed makes sense. I see it visually what I would like to accomplish but I may not be articulating it correctly.

Suppose there is a different limit point $\mathbf b$ of $\mathbf B$. Set $\varepsilon=\frac12|\mathbf b-\mathbf a|$. Then by convergence of the sequence there are only finitely many points outside $B_ε(\mathbf a)$. 
However at the same time, as $\mathbf b$ is a limit point, there need to be infinitely many points of $\mathbf B$ inside $B_ε(\mathbf b)$, which is contained in that "outside". 
This is a contradiction.
