How to give a rigorous proof of this fact about closures of open balls in the euclidean spaces? Let $n$ be a positive integer, $\vec{a} \in \mathbb{R}^n$, and $r > 0$. Then it is intuitively clear that the closuer of the open ball $$B(\vec{a} ; r) \colon= \{ \vec{x} \in \mathbb{R}^n \colon ||\vec{x} - \vec{a} || < r \}$$ is the closed ball $$\tilde{B}(\vec{a}; r) \colon= \{ \vec{x} \in \mathbb{R}^n \colon ||\vec{x} - \vec{a} || \leq r \}. $$ How to give a rigorous, yet elementary, proof of this? 
By the closure $\bar{S}$ of a set $S$, we mean the union of the set $S$ with the set $S^\prime$ of all the limit points of $S$. 
 A: Suppose $x_n$ is a sequence in $B(a,r)$ that converges to $y$. Then, you can use
$$
|y-a|\leq |y-x_n|+|x_n-a|<|y-x_n|+r
$$
and the fact that the weak inequality is preserved under the limit to show that $\bar{B}(a,r)$ contains all limit points of $B(a,r)$.
For the reverse inclusion, suppose $y\in\bar{B}(a,r)$ and write $v=y-a$. Then let $x_n=\frac{n}{n+1}v+a$ so that $x_n\in B(a,r)$ and $x_n\to y$. 
A: As usual, to prove two sets are equal, show they are subsets of each other.  For notational ease, let the open ball be called B and the closed ball be called $B^*$.   Let $x \in cl(B)$.  Then $x \in B$ or $x \in B'$.  If $x \in B$, then $X \in B^*$ and we are done Let $x\in B'$.  Then for every neighborhood A, $x\in A$,  $A \cap B \ne \emptyset$.  In particular, this is true for every open ball of radius $\epsilon$ around x.  Now, measure the distance from a to x.  We have some point y in the intersection of B and A,  so by the triangle inequality, we have $d(a,x)\le d(a,y)+d(y,x)$.   Since $y\in B$,  we have $d(a,y)<r$.  We have $d(y,x)<\epsilon$.   So, for any epsilon, we have $d(a,x)<r+\epsilon $,  hence $d(a,x)\le r$,  hence $x\in B^*$.
Conversely, let $x\in B^*$.   Then $d(x,b)\le r$.  Then $d(x,b)<r$,  in which case $x\in B$,  or $d(x,b)=r$.  If the latter,  for any $\epsilon$ neighborhood of x, we have a point y in that neighborhood such that $y\ne x$  and $d(y,x)<r$,  hence $y\in B'$,  hence $y\in cl(B)$.  Thus the two sets are equal
A: Let $y \in \Bbb{R}^n$ such that $|a-y|=r$. Now let $V \subset \Bbb{R}^n$ be any open set such that $y \in V$. Since $V$ is open we can place an open ball inside of $V$ centered at $y$ for some radius $\epsilon>0$. That is, $B(y,\epsilon) \subset V$. If $\epsilon \geq r$ then we know $B(a,r) \cap [B(y,\epsilon) -\{ y\}] \neq \emptyset$ because $\frac{a+r}{2} \in B(a,r) \cap [B(y,\epsilon) -\{ y\}]$. So suppose $\epsilon<r$. Then we know $r>r-\frac{\epsilon}{2}>r-\epsilon>0$, so we know $r-\frac{\epsilon}{2} \in B(a,r) \cap [B(y,\epsilon)-\{y\}]$. Now that we know the set $B(a,r) \cap [B(y,\epsilon)-\{y\}]$ is non-empty regardless of the size of $\epsilon$ and that $B(y,\epsilon) \subset V$ of any arbitrary open set $V$, we know that $B(a,r) \cap [V-\{ y\}] \neq \emptyset$ for all $y$ where $|a-y|=r$. This means the set $\mathcal{P}=\{y \in \Bbb{R}^n:|a-y|=r \}$ is a collection of limit points not contained in $B(a,r)$. Now I will leave it up to you to prove that $p$ is not a limit point of $B(a,r)$ if $|a-p|>r$. This can be done by choosing a small enough radius in an open ball around $p$, and showing it will not intersect $B(a,r)$. Once you know that there are no limit points outside of $B(a,r)$ besides those defined in $\mathcal{C}$, it follows that $$B(a,r) \cup \mathcal{C} = \overline{B(a,r)} = \{x \in \Bbb{R}^n: |a-x| \leq r\}$$
A: Noting that the complement of $\overline{B(a,r)}$ is open, we see that $\overline{B(a,r)}$ is closed and hence must contain the closure of $B(a,r)$. This can be seen by putting a sufficiently small open ball around every point so that they do not intersect with $\overline{B(a,r)}$.
To see that $\overline{B(a,r)}$ is in fact the closure is simple, we now just need to show that all points with $||x-a||=r$ are limit points. To see this fix $x$ at a distance $r$ from $a$ then for any $r>\epsilon>0$ consider the point $\frac\epsilon r(a-x)+x$. It should be clear by construction that this point is in the ball and can be used to show $x$ is a limit point.
Explicitly, the point is in $B(a,r)$ as $$||a-\frac\epsilon r(a-x)-x||=(1-\frac\epsilon r)||a-x||=(1-\frac\epsilon r)r<r.$$
In addition the distance to $x$ is given by
$$||x-\frac\epsilon r(a-x)-x||=\frac\epsilon r||a-x||=\epsilon$$
and so $x$ is a limit point.
