If $x \in \{f \leq a\}$, you have that $f(x) \leq a$. So $f(x) \leq a + \frac{1}{k}$ for all $k$. It has been shown that $\{f \leq a\} \subset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$.
If $x \in \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$. Then $f(x) < a + \frac{1}{k}$ for all $k$. This can only happen if $f(x) \leq a$. So $\{f \leq a\} \supset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$
The first equality of sets has been shown.
The second is similar.
If $x \in \{f < a\}$, then $a - f(x) > 0$. Choose a $k$ such that $\frac{1}{k} \leq a - f(x)$. Then $f(x) \leq a - \frac{1}{k}$. Hence $x \in \{f \leq a - \frac{1}{k} \}$. Thus, $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$.
Suppose $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$. This means that $f(x) \leq a - \frac{1}{k}$ for some $k$. So $f(x) < a$. $x \in \{f < a\}$.
Equality of the two sets has been shown.