Am I missing something quite obvious from understanding this proof? 
Theorem. There exists a nonempty set of rational numbers which is bounded above in $\mathbb{Q}$ but has no least upper bound in $\mathbb{Q}$.

Here is the relevant part of the proof that I am going to ask about.

Let $k=\frac{a}{b}\in\mathbb{Q}$ be an upper bound for the set $S=\{q\in\mathbb{Q}:q^2<2\}$. Suppose that $k^2<2$. Define $\delta=2-k^2>0$. Fix $N\in\mathbb{N}$ such that
  $$
N\geq\max\{2a+1,3a/(b^2\delta)\}
$$
  Since $N\geq 2a+1$, we have $N^2\geq N(2a+1)\geq 2Na+1$ and since $N>3a/(b^2\delta)$, then $N^2b^2\delta>3Na\geq 2Na+1$ which forces
  $$
\frac{2Na+1}{N^2b^2}<\delta
$$
  Define $l=\frac{Na+1}{Nb}$. Then
  $$
l^2=k^2+\frac{2Na+1}{N^2b^2}<k^2+\delta=2
$$

What is my question?
No matter how hard I think about this part of the proof I see nowhere that $N\geq 2a+1$ is used to show that $l^2<2$. I am sure that I am missing something quite obvious but I can not figure it out. Perhaps the thinking of the author of the proof was different to mine. Why was it necessary to fix $N\in\mathbb{N}$ such that
$$
N\geq\max\{2a+1,3a/(b^2\delta)\}
$$
and not just $N> 3a/(b^2\delta)$? In other words, I see $N\geq 2a+1$ unnecessary.
$k=\frac{a}{b}\in\mathbb{Q}$ is an upper bound in $\mathbb{Q}$ for $S$. We know that $\sqrt{2}\approx1.41421356237$ is the supremum of $S$ in real numbers so $k\geq 1.5$ or $k\geq 1.42$ or $k\geq 1.415$ and so on which forces $a>1$.
$l^2=k^2+\frac{2Na+1}{N^2b^2}$ and I can approximate $\frac{2Na+1}{N^2b^2}$ as follows
$$
\begin{align*} \frac{2Na+1}{N^2b^2}&\leq\frac{2Na+N}{N^2b^2}&&\text{since }N\geq 1\\ &=\frac{2a+1}{Nb^2}&&\\ &<\frac{2a+a}{Nb^2}&&\text{since }a>1\\ &=\frac{3a}{Nb^2}&& \end{align*}
$$
So I need $\frac{3a}{Nb^2}<\delta$ which implies $N>\frac{3a}{b^2\delta}$.
 A: 
The author presents a proof purely based upon calculations within $\mathbb{Q}$. It does not rely on an embedding of $\mathbb{Q}$ in $\mathbb{R}$ and the knowledge that $\mathbb{R}$ is complete.
Insofar is the approach of the author somewhat different than yours.

Let's revisit the author's proof. We want to show there does not exist a rational number $k$ which is a least upper bound of $S$, the set of rational numbers having square less than $2$.
In order to do so we show that whenever we want to take a least upper bound $k=\frac{a}{b}\in\mathbb{Q}$ of $S$ with $k^2<2$, we find a rational number $l$ in $S$ with $l>k$ contradicting the assumption $k$ is a least upper bound.

We consider the distance $\delta=2-k^2=2-\frac{a^2}{b^2}$ from the square of $k$ to $2$ and construct a rational number $l\in S$ having a smaller distance. 
We take a natural number $N$ with
  \begin{align*}
N\geq\max\{2a+1,3a/(b^2\delta)\}
\end{align*}
Since $N\geq 2a+1$ and since $N>\frac{3a}{b^2\delta}=\frac{3a}{b^2\left(2-\frac{a^2}{b^2}\right)}=\frac{3a}{2b^2-a^2}$ we obtain
  \begin{align*}
N^2b^2\delta&=N^2b^2\left(2-\frac{a^2}{b^2}\right)=N^2(2b^2-a^2)\\
&>3Na\\
&\geq 2Na+1
\end{align*}
  which implies
  \begin{align*}
\frac{2Na+1}{N^2b^2}<\delta
\end{align*}
If we now define $l:=\frac{Na+1}{Nb}>\frac{Na}{Nb}=k$ we obtain
  \begin{align*}
l^2&=\frac{(Na+1)^2}{N^2b^2}=\frac{N^2a^2+2Na+1}{N^2b^2}\\
&=\frac{a^2}{b^2}+\frac{2Na+1}{N^2b^2}\\
&<\frac{a^2}{b^2}+\delta\\
&=2
\end{align*}
and the claim follows.

Note: In accordance with OP's comment below we don't need the following part of the author's proof:
\begin{align*}
N^2\geq N(2a+1)=2Na+N>2Na+1
\end{align*}
