A compact subset, E, of $\mathbb{R^n}$ is bounded. Let $E\subset\mathbb{R^n}$ be a compact set. Let $$B(\mathbf{x},1)=\{\mathbf{y}\in\mathbb{R^n} : |\mathbf{y}-\mathbf{x}|<1\}.$$ Let $\{B(\mathbf{x},1)\}_{\mathbf{x}\in E}$ be an open cover of $E$. This is indeed an open cover of $E$ since $$\mathbf{x}\in\{B(\mathbf{x},1)\}_{\mathbf{x}\in E}$$ for all $\mathbf{x}\in E$.  Since $E$ is compact, there exists a finite sub-cover of $E$, namely $\{B(\mathbf{x}_i,1)\}_{i=1}^n$. Let $D=\max\{|\mathbf{x}_1-\mathbf{x}_j|\}$ where $1\leq j\leq n$. This maximum is defined because the set of $\{\mathbf{x_1},\mathbf{x_2},...,\mathbf{x_n}\}$ is finite.
I'm now struggling in showing how $E\subset B(\mathbf{x},D)$ in order to conclude that $E$ is bounded.
 A: Rather, the thing you want to show is that $E \subset B(x_1, D\!+\!1)$.  The problem is that $D$ is only the max distance between the centers of the balls, not their boundaries, so $E$ is not necessarily contained inside of $B(x_1, D)$.  Perhaps the easiest way to see this is to just draw a picture with two disjoint balls.  Now, given any $z \in E$, we want to show that $z \in B(x_1, D\!+\!1)$.  We'll have $z \in B(x_k, 1)$ for some $k$, so applying the triangle inequality:
$$d(x_1,z) \leq \underbrace{d(x_1, x_k)}_{\leq D} + \underbrace{d(x_k, z)}_{< 1} <D + 1$$
Indeed, $z \in B(x_1, D+1) \implies E \subset B(x_1, D+1)$.

To avoid this headache, there's a simpler open cover: for any $y \in E$, consider $\{B(y, n) \}_{n=1}^\infty$.  Because these open sets are "nested" in the sense that $B(y, n) \subset B(y, n\!+\!1)$ for all $n \in \mathbb{N}$, the finite subcover that this open cover admits will simply be $B(y, m)$ for some $m \in \mathbb{N}$: clearly bounded.
A: As carmichael noted, you probably are aiming to show $E \subset B(x_1, D+1)$.
Hint: you want to show that there is some finite number $C$ such that for any $x \in E$, we have $|x - x_1|<C$. Use your finite set $\{x_1,\ldots,x_n\}$ and the triangle inequality.
Hint:

 for each $x \in E$ there exists some $x_i$ in your finite set such that $|x-x_i| < 1$.

Hint: then, continuing from the previous hint,

 $$|x-x_1| \le |x-x_i| + |x_i - x_1| < 1 + D$$

