Showing that $A=\{a_1,...,a_r\}$ is a closed set Let $A=\{a_1,\dots,a_r\}, a_i \in \mathbb{R}, i=1,...,r$. Show that $A$ is a closed set. As a bit of a beginner, I have written down a proof and I wanted to see if it is good enough/well structured.
So I want to show that $\partial A\subseteq A$, where $\partial A=\{a \in A:D(a,\delta) \cap A\neq\emptyset, D(a,\delta)\cap(\mathbb{R}\setminus A)\neq\emptyset, \forall\delta>0\}$ is the set of boundary points of $A$. The neighbourhood of $A$ for any $\delta>0$ is denoted by $D(a,\delta)=(a-\delta,a+\delta)$.
Let $a \in A$. For all $\delta>0$, $D(a,\delta)\cap A\neq\emptyset$ (since $D(a,\delta)\cap A=\{a\}$ even for arbitrarily small $\delta$). Also, $D(a,\delta)\cap(\mathbb{R}\setminus A)\neq\emptyset$.  So $a \in \partial A$ for all $a \in A$.
Now let $B=\mathbb{R}\setminus A$ and $y\in B$. If $y<\min A$, then $D(y, \frac{\min A-y}{2})\subseteq B$, so $y$ is exterior to $A$. If $y>\max A$, then $D(y,\frac{y-\max A}{2})\subseteq B$, so $y$ is exterior to $A$. If $a_j<y<a_k$, where $a_j\in A$ and $a_k=\min(A\setminus\{a\in A: a\leq a_j\})$, and $\delta=\frac{1}{2}\min(y-a_j,a_k-y)$, then $D(y,\delta)\subseteq B$, so $y$ is exterior to $A$. Therefore every element in $B=\mathbb{R}\setminus A$ is exterior to $A$. 
So $\partial A=\{a_1,\dots,a_r\}=A$, or $\partial A\subseteq A$, so $A$ is a closed set.
1) Is my conclusion (and the way I arrived at it) correct? Is every element in $A$ a boundary point of $A$?
2) I think I've made my proof a bit more complicated than it should be, is there a simpler way?
3) Tips for formatting/notation?

It seems to me the introduction of $\partial A$ into the discussion overly complicates things.  
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 \le k \le r$, the set $\Bbb R \setminus \{ a_k \}$ is open; this is easy to see, since if $p \in \Bbb R \setminus \{a_k\}$, the open interval
$(p - \delta, p + \delta) \subsetneq \Bbb R \setminus \{a_k\}, \tag 1$
where
$\delta = \dfrac{\vert p - a_k \vert}{2}, \tag 2$
contains $p$ and is fully contained in $\Bbb R \setminus \{a_k \}$; thus $\Bbb R \setminus \{ a_k \}$ is open, since it contains an open neighborhood $(p - \delta, p + \delta)$ of any its points $p$; thus the complement of $\Bbb R \setminus \{a_k\}$, which is the singleton $\{a_k\}$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = \{ a_1, a_2, \ldots, a_r \} = \displaystyle \bigcup_{k = 1}^r \{ a_k \}. \tag 3$
