Just using unions of intervals of the form $(x,y]$ will never give you an interval of the form $[a,b)$:

Suppose $[a,b) = \bigcup_{n \in \mathbb{N}} (a_n,b_n]$ for some $a_n,b_n \in \mathbb{R}$. Because $a \in [a,b)$ there exists $n \in \mathbb{N}$ such that $a \in (a_n,b_n]$, hence there exists $\varepsilon > 0$ such that $a - \varepsilon \in (a_n,b_n]$. This implies that $a - \varepsilon \in [a,b)$ which is a contradiction.

The same argument can be used to prove that intervals of the form $[a,b]$ cannot be written as unions of intervals of the form $(x,y]$.

Edit:

Concerning intersections, you have $[a,b] = \bigcap_{n \in \mathbb{N}} (a-\frac{1}{n},b]$.

Suppose $[a,b) = \bigcap_{n \in \mathbb{N}} (a_n,b_n]$. As $b \notin [a,b)$ there exists a $n \in \mathbb{N}$ such that $a \leq b_n < b$. This implies that there exists an $\varepsilon > 0$ such that $b_n + \varepsilon < b$. But then we have $b_n + \varepsilon \in [a,b)$ which contradicts $b_n + \varepsilon \notin (a_n,b_n]$.

So $[a,b)$ cannot be represented as the intersection of intervals of the form $(x,y]$, either. Of course, combining intersections and unions will let you succeed as $[a,b) = [a,c] \cup (c,b)$ for $c \in [a,b)$ and for these you have such representations (this was already mentioned in the comments).

In case this is a homework, here is a hint on a similar example.

Let $$ A = \{ x \in \mathbb{R} | x^2-x-6 > 0 \}.$$ First, factor $ x^2-x-6 = (x-3)(x+2) = 0$. Hence there are two roots at $x =3, x= -2$. This divides up the number line $\mathbb{R}$ into three intervals: $$ (-\infty, -2), (-2, 3), (3, \infty). $$ Now you need to check the sign of $(x-3)(x+2)$ in each of the three intervals. Only $x \in (3, \infty) \cup (-\infty, -2)$ makes $(x-3)(x+2)$ positive, i.e. $x^2-x-6 > 0.$ Hence $$ A = (3, \infty) \cup (-\infty, -2). $$ Do the same with $A,B$ in your question. Then $A \cup B$ should be apparent. Note: pay attention that $B$ has $\le$ in the definition, so some intervals will be closed (rather than open).

Let's think formally about what a boundary is. If you have a set $A$, with closure $\bar{A}$ and interior $\mathring{A}$, then the boundary of $A$ is $\partial{A} = \bar{A} \setminus \mathring{A}$.

Let $A = \{1,2,3\} \cup (2,4)$.

What is the closure of this set? The easy way is to find the points whose neighborhoods always contain some points in $A$ (the closure is the set of these points, by definition). In this case, the closure is $\{1\} \cup [2,4]$.

The interior is, by definition, the set of points who have at least one neighborhood contained totally in the set. So, the interior of this set is $(2,4)$.

The boundary of the set is therefore $\bar{A} \setminus \mathring{A} = \{1,2,4\}$, just three points!

You can always display a set in $\mathbb{R}$ as a union of points (singletons) with intervals. So, the original set would be $\{1\} \cup [2,4)$ as you essentially suggested.

Sort them by one of the terms (start, for example), then check for overlaps with its (right-hand) neighbor as you move through the list.

class tp:
    def __repr__(self):
        return "(%d,%d)" % (self.start, self.end)

    def __init__(self, start, end):
        self.start = start
        self.end = end


s = [tp(5, 10), tp(7, 8), tp(0, 5)]
s.sort(key=lambda self: self.start)
y = [s[0]]
for x in s[1:]:
    if y[-1].end < x.start:
        y.append(x)
    elif y[-1].end == x.start:
        y[-1].end = x.end

You could write the union as $\{1,3\}\cup [4,\infty)$ but it does not simplify more. Espacially you can not write this as an interval, what you might want.

Keep in mind that $(4,\infty)=\{x\in\mathbb{R}| 4<x\}$