Note well that the only subsets of $\mathbb R^n$ ($n\in\mathbb N$) that are both closed and open are the whole set and the empty set. (This is the consequence of a fundamental property of Euclidean spaces called connectedness, which is the topological formalization of the intuitive geometric notion of contiguity.) Since $\mathbb R^n$ is not compact (it's unbounded), the only compact open subset in this space is the empty set.
Nevertheless, a bounded set may very well be open. Just consider any open interval $(a,b)$ ($a\in\mathbb R$, $b\in\mathbb R$, $a<b$) for $n=1$, any circle without its perimeter for $n=2$, or any sphere without its surface for $n=3$.
At any rate, if one digresses from the usual Euclidean topology on $\mathbb R^n$, interesting things can happen. Endow $\mathbb R^n$ with the discrete topology, in which every subset is declared to be open. But then the complement of every subset is open too, so every subset is closed as well. Consequently, every subset of $\mathbb R^n$ is both open and closed in the discrete topology! Moreover, one can show that a subset is compact with respect to this topology if and only if it contains only finitely many points, so the Heine–Borel theorem (formulated for the usual Euclidean topology) no longer applies to the discrete topology!
For the finite union case, we will show that if $E_1 \subseteq X$ and $E_2 \subseteq X$ are bounded, then so is $E_1 \cup E_2$. That finite unions of bounded sets is bounded then follows by induction.
Since $E_1$ is bounded, there is some $x_1 \in X$ and $r_1 \in \mathbb{R}^+$ such that $E \subseteq B(x_1, r_1)$. Since $E_2$ is bounded, wer know that there is some $x_2 \in X$ and $r_2 \in \mathbb{R}^+$ such that $E_2 \subseteq B(x_2, r_2)$.
Let $r = r_1 + r_2 + d(x_1, x_2)$. Then if for any $x \in E_1 \cup E_2$, we have that either $x \in E_1$ or $x \in E_2$. If $x \in E_1$, then we know that $d(x, x_1) < r_1 < r$. If $x \in E_2$, then we have that $d(x, x_1) \leq d(x, x_2) + d(x_2, x_1) < r_2 + d(x_1, x_2) < r$ by the triangle inequality. Thus we have that $x \in B(x_1, r)$.
We see that $E_1 \cup E_2 \subseteq B(x_1, r)$, and hence $E_1 \cup E_2$ is bounded.
For the arbitrary intersection case, suppose that $\{E_i\}_{i \in I}$ is an arbitrary non-empty family of bounded sets. Let $E$ be their intersection. Then for any $i \in I$, we have that $B_i$ is bounded, and hence there is $x \in X$ and $r \in \mathbb{R}^+$ such that $E_i \subseteq B(x, r)$. But $E \subseteq B_i \subseteq B(x, r)$, and hence $E$ is also bounded.
Essentially, if you pick any set out of those that you're taking the intersection of, the intersection will be contained in that set. Since that set is bounded by assumption, so is the intersection.
Best Answer
Let ${\rm dist}\,(A,B) = \inf \{d(x,y): x \in A, y \in B\}$ denote the distance between sets.
Suppose $A_0$ and $B_0$ are closed bounded nonempty sets in ${\mathbb R}^n$ with $d(A_0,B_0) = 0$. $A_0$ is the union of finitely many closed bounded nonempty sets of diameter at most $1$ (e.g. take the intersections with rectangles of diameter at most $1$ that tile a large rectangle enclosing $A_0$), and at least one of these (call it $A_1$) must have ${\rm dist}\,(A_1, B_0) = 0$. Similarly find $B_1 \subset B_0$ of diameter at most 1 such that ${\rm dist}\,(A_1, B_1) = 0$.
Repeating this idea, inductively define sequences of closed, bounded nonempty sets $A_n$ and $B_n$ with $A_{n+1} \subset A_n$, $B_{n+1} \subset B_n$, such that ${\rm dist}\,(A_n, B_n) = 0$ and $A_n$ and $B_n$ have diameter at most $1/n$. If $x_n \in A_n$ and $y_n \in B_n$, then $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences and converge respectively to $x \in A_0$ and $y \in B_0$, and it's easy to see that $d(x,y) = 0$, so $x = y$. Thus $A_0$ and $B_0$ are not disjoint.