Remember that if we have a $\sigma$-algebra, then any countable intersection of sets in the $\sigma$-algebra must also be in the $\sigma$-algebra.
Also, note that any closed interval on $\mathbb{R}$ is of the form $(-\infty,a]$ or $[a,b]$ or $[b,\infty)$ or $(-\infty,\infty)$, where $a,b \in \mathbb{R}$.
$(-\infty,\infty)$ is also open and hence is in the $\sigma$-algebra.
Now lets consider $[a,b]$. The important thing to recognize is that $[a,b]$, where $a,b \in \mathbb{R}$, can be written as a countable intersection of open intervals, for instance $[a,b] = \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$
Since each of $\displaystyle \left(a - \frac1n, b+ \frac1n \right) \in \sigma$-algebra, so does the countable intersection which is nothing but $[a,b]$.
Closed intervals of the form $(-\infty,a]$ can be written as $\displaystyle \bigcap_{n=1}^{\infty} \left(-\infty, a + \frac1n \right)$ and similarly, closed intervals of the form $[b, \infty)$ can be written as $\displaystyle \bigcap_{n=1}^{\infty} \left(b - \frac1n , \infty \right)$.
EDIT:
To prove, $\displaystyle [a,b] = \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$, we will prove that $[a,b] \subseteq \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$ and $$[a,b] \supseteq \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right).$$
Consider $x \in [a,b]$. This means that $a \leq x \leq b$. However, $a - \frac1n < a$ and $b < b + \frac1n$, for all $n \in \mathbb{Z}^+$. Hence, we have that $a - \frac1n < a \leq x \leq b < b + \frac1n,$ for all $n \in \mathbb{Z}^+$. Hence, $\displaystyle x \in \left(a - \frac1n, b+ \frac1n \right)$, forall $n \in \mathbb{Z}^+$. Hence, $$x \in \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right).$$
Hence, $$[a,b] \subseteq \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right).$$
Now consider $\displaystyle x \in \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$. This means that $a - \frac1n < x < b + \frac1n,$ forall $n \in \mathbb{Z}^+$. If $x \notin [a,b]$, say $x < a$, by Archimedean property, we can find a $m \in \mathbb{Z}^+$ such that $x < a-\frac1m < a$. But this means that $\displaystyle x \notin \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$. Similarly, if $x > b$, then by Archimedean property, we can find a $m \in \mathbb{Z}^+$ such that $ x > b + \frac1m $. But this means that $\displaystyle x \notin \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right)$. Hence, $x \in [a,b]$. Hence, $$[a,b] \supseteq \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right).$$
From the above, we can hence conclude that $$[a,b] = \bigcap_{n=1}^{\infty} \left(a - \frac1n, b+ \frac1n \right).$$