Set $B$ is the boundary of $R$ which is a subset of $\mathbb{R}^2$. If $B$ is bounded, closed, and compact, show that a given set is an open cover. Let $R \subset \mathbb{R}^2$ be a bounded set (you might say a bounded region in the plane).  Let
$B=\partial R$ be its boundary.  Show that for any $\delta>0$, $B$ can be covered by a finite number of disks of radius $\delta$ with centers in $B$. That is, there are $x_i\in B$ for $i=1,\ldots,n$ so that $\{B_{<\delta}(x_i)\mid i=1,\ldots,n\}$ is an open cover of $B$. 
I have already proved that $B$ is bounded and compact, so I know that is satisfies the condition where every open cover has a finite subcover.  I am not entirely sure how to prove this specific set is an open cover.  I was debating if I should use the distance between points in the set and show that any open ball around points will overlap and cover the whole set.  I am new to real analysis and am not even sure if my thinking is correct or how to go about proving this.  Any help would be greatly appreciated.
 A: For any $x\in\partial B$, There exsit a sequence $\{x_n\}\subseteq B^0$ such that $x_n\to x$. Therefore, for given $\delta$, there exist $N=N(x)$ such that $d(x_N,x)\leq \delta/2<\delta$. Let $B_x=B_\delta(x_N)$. Then $\{B_x:x\in \partial B\}$ is an open cover. 
Since $B$ is compact, it must have a finite subcover. 
Alternatively, just consider the cover $\{B_\delta(x):x\in B^0\}$.
$\partial B$ is compact because it is a closed subset of a compact set $\overline B$.
A: Consider the set of all discs with radius $\delta$ centered at points of $B$. They form an open cover of $B$ and, since $B$ is compact (since it is closed and bounded), this open cover of $B$ has a finite subcover.
A: We can even choose the balls so that the centers are in $R$.
Fix $\delta > 0$. The family $\left\{B\left(x, \frac{\delta}2 \right): x \in B\right\}$ is an open cover for $B$. Since $B$ is compact, there exist $x_1, \ldots, x_n \in B$ such that $\left\{B\left(x_1, \frac{\delta}2 \right), \ldots, B\left(x_n, \frac{\delta}2 \right)\right\}$ covers $B$.
Since $x_i \in B = \partial R$ there exists $y_i \in R$ such that $d(x_i, y_i) < \frac\delta2$. Then $$B\left(x_i, \frac\delta2\right) \subseteq B\left(y_i, \delta\right)$$
so $\{B(y_1, \delta), \ldots, B(y_n, \delta)\}$ covers $B$.
