Terence Tao, Analysis I, Thm. 5.5.9, Least upper bound Terence Tao, Analysis I, 3e, Thm. 5.5.9:

Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty
  subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e., $E$ has some
  upper bound $M$), then it must have exactly one least upper bound.

In the proof it says that

(...) Now we show it [$S := \text{LIM}_{n \rightarrow \infty} (m_n - 1)/n$] is a least upper bound. Suppose $y$ is an
  upper bound for $E$. Since $(m_n - 1)/n$ is not an upper bound, we
  conclude that $y \ge (m_n - 1)/n$ for all $n \ge 1$.

But from  my understanding, if $y$ is an upper bound for $E$, then all elements in $E$ are smaller than or equal to $y$. And since $(m_n - 1)/n$ is not an upper bound, there are elements $x$ in $E$ such that 
$$
(m_n - 1)/n < x \le y.
$$
If $y > (m_n - 1)/n$, then $y \ge (m_n - 1)/n$.
But why is equality used here? Is it some sort of trick to make use of 

(...) if $a_n \le x$ for all $n \ge 1$, then $\text{LIM}_{n
\rightarrow \infty} \; a_n \le x$

in a following step of the proof?
 A: All that can be said is that $(m_n -1)/n \leq y$; that is by definition. Given the fact that $\forall n \in \mathbb{N}$ the element $(m_n -1)/n$ is not an upper bound we can deduce that there is strict inequality.
In the following sentence he uses Exercise 5.4.8, which essentially says 

(...) if $a_n \le x$ for all $n \ge 1$, then $\text{LIM}_{n
\rightarrow \infty} \; a_n \le x$

Here the result is true regardless of whether $a_n \le x$ or $a_n < x$ for all $n \ge 1$. In short, it does not matter whether to write strict or non-strict inequality; both are valid and the conclusion is the same.

By the way 

if $a_n < x$ for all $n \ge 1$, then $\text{LIM}_{n
\rightarrow \infty} \; a_n < x$

is NOT true. Consider the sequence $(\frac{-1}{n})_{n \in \mathbb{N}}$. Here each element is strictly smaller than $0$, but the limit equals $0$.

Also, the statement 

And since $(m_n - 1)/n$ is not an upper bound, there are elements $x$ in $E$ such that 
  $$ (m_n - 1)/n < x \le y.$$

is only true because $\mathbb{R}$ is totally ordered. In a general partially ordered set this is not nessecarily the case.
