Need help on understanding the theorem in real analysis! 
Theorem 1.3.3: Let $L \in R$ and let $x_n$ be a sequence of real
  numbers. Then $x_n \to L$ if and only if $\limsup x_n =
 \liminf x_n = L$.

Proof: First, suppose $x_n \rightarrow L.$ Thus if $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $ n \geq N$ implies $|x_n-L| < \epsilon/2$, which implies $s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$. 
Thus
$$L-\frac{\epsilon}{2} \leq i_n \leq s_n \leq L +\frac{\epsilon}{2}$$
which implies that $|s_n-L| \leq \frac{\epsilon}{2} \lt \epsilon$ and $|i_n-L| \leq \frac{\epsilon}{2}<\epsilon$, for all $\epsilon > 0$. Thus $s_n \rightarrow L $ = lim sup $x_n$ and $i_n \rightarrow L$ = lim inf $x_n$.
I only included the first part of the proof. I boxed the part where I was confused. Why did they use  inequalities 
$s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$ instead of 
$s_n = \sup(T_n) \lt L + \epsilon/2$ and $i_n = \inf(T_n) \gt L-\epsilon/2$. 

There is a reason to not use strict inequalities. For arbitrary $S \subseteq \mathbb{R}$, it could happen that $x < L$ for all $x \in S$, but $\sup S = L$. (but never $\sup S > L$)
For example, take $S = \left]0,1\right[$ and $L = 1$.
However, since $x \to L$ is assumed, you can safely replace $\leq$ with $<$ since the limiting case $\sup T_n = L + \epsilon/2$ never happens.
