Understanding why the author chose the number he did in this proof that $\sqrt 2$ exists

This is just goal-oriented work: The author wants $$\tag1\left(\alpha+\frac1n\right)^2<2.$$ Equivalently, $$\alpha^2+\frac{2\alpha}n+\frac1{n^2}<2. $$ Rearrange to $$\frac{2\alpha}n+\frac1{n^2}<2-\alpha^2. $$ This looks promising because, by assumption, the right hand side is positive and we only have to find $n$ that makes the left (positive and) small enough. As $\frac1n\le\frac1{n^2}$, we can boldly strengthen our task to find $n$ that makes (even) $$\frac{2\alpha}n+\frac1{n}<2-\alpha^2. $$ (The advantage of this is that the dependence on $n$ is now simpler than with the square part).

Now multiply with the positive(!) $n$ $$ 2\alpha+1<(2-\alpha^2)n$$ and divide by the positive(!) $2-\alpha^2$ to arrive at $$ \frac{2\alpha+1}{2-\alpha^2}<n$$ as a sufficient (and readily fulfilled) condition for $n$ to make $(1)$ true.

The author is trying to say there exists an sufficiently large $n_0$ s.t. $\alpha^2 + \frac{2 \alpha + 1}{n_0} < 2$, which

$\iff \frac{2 \alpha + 1}{n_0} < 2-\alpha^2$

$\iff \frac{2 \alpha + 1}{n_0} < 2-\alpha^2$

$\iff \frac{1}{n_0} < \frac{2-\alpha^2}{2\alpha-1}$