Proof for $\epsilon$ closeness to $\sqrt{2}$ for all values of $\epsilon$

I am trying to understand the proof above, Proposition 7, given by the author, but I got stuck at some points.

My explanation:

Now, suppose for contradiction there is an $\epsilon > 0$, for all $x > 0$ such that $NOT (x^2 < 2 < (x+ \epsilon)^2)$,

Since $NOT (x^2 < 2 < (x+ \epsilon)^2) \ \ iff \ \ (x^2 \geq 2 \ \ OR \ \ (x+ \epsilon)^2 \leq 2)$,

We must show that for some $\epsilon>0$ and for all $x > 0$, the statement $x^2 \geq 2 \ \ OR \ \ (x+ \epsilon)^2 \leq 2$ leads to a contradiction.

I think in the proof above, author tries to get a contradiction using right hand side of $OR$ clause, namely $(x+ \epsilon)^2 \leq 2$. But no contradicition is achieved by him for the left hand side side of $OR$ clause namely, $x^2 \geq 2$.

First Question is, am I correct at this point, i.e. the need for a contradiction for the left hand side side of $OR$ clause?

Second question is, I don't understand how he derived the following sentence in his proof above,

Since $0^2 < 2$, we thus have $\epsilon^2 < 2$, which then implies $(2 \epsilon)^2 < 2$ and indeed $(n \epsilon)^2 < 2$

Why does he feel a need to multiply $\epsilon$ constantly by $n$

Answer to the first question: you are kind of wrong. I understand your logic, but it is not needed in the authors argument. You are forgetting the fact that, in this case, $(x+\epsilon)^2>x^2$. So the negation of the original statement is that $2$ is either on one side or the other of both numbers.

The author shows that, starting from that OR statement, a contradiction is achieved. He doesn't need to "use" the other side. Already taking $x=0$ leads to a contradiction.

Answer to the second question: He doesn't "feel the need"; it is just a reasoning that leads to the desired contradiction. The point is: he takes $x=0$. Then $0^2<x^2$. This implies that $(0+\epsilon)^2<2$. So now we know that $\epsilon^2<2$. So now you can take this as your $x$, and conclude that $(\epsilon+\epsilon)^2<2$. Now you take $2\epsilon$ as your $x$ (as we just deduced that $(2\epsilon)^2<2$), to get $(2\epsilon+\epsilon)^2<2$. As the author says, one can continue this way to conclude that $n^2\epsilon^2=(n\epsilon)^2<2$ for all $n$, so $n^2<2/\epsilon^2$ for all $n$. This is the desired contradiction, as the set of natural numbers is not bounded above.

The proof is not for some $\epsilon$, but rather for any positve rational $\epsilon$. The author does not exactly what you write. Rather, he uses the the negation of the claim: With $\epsilon>0$ given, the claim of proposition 7 is:

"There exists a nonnegative number $x$ such that $x^2<2$ and $2<(x+\epsilon)^2$."

The negation of this is:

"For every nonnegative $x$, we have $x^2\ge2$ or $2\ge(x+\epsilon)^2$."

Thus if (under this assumption) $x$ is a nonnegative number with $x^2<2$, the first alternative is wrong, hence the second must be true (and by irrationality of $\sqrt 2$ we can replace $\ge$ with $>$ in it). In short, the author uses specifically $$\tag1 x^2<2\Rightarrow (x+\epsilon)^2<2 \qquad\text{if }x\ge0,$$ to arrive at a contradiction with the archimedean property. To do so, he shows by induction that $(n\epsilon)^2<2$ holds for all $n\in \mathbb N$. The case $n=0$ is trivial as $(0\epsilon)^2=0<2$. The induction step follows using (1): If $(n\epsilon)^2<2$ apply (1) with $x=n\epsilon$ to find $$((n+1)\epsilon)^2=(n\epsilon+\epsilon)^2=(x+\epsilon)^2<2.$$ Since for sufficiently big $n$ we have $n\epsilon\ge2$ (and hence ($(n\epsilon)^2\ge4>2$), we arrive at a contradiction. Since $\epsilon$ is rational, say $\epsilon=\frac ab$ with $a,b\in\mathbb N$, we can in fact be very specific: For $n=2b$ we have $n\epsilon = 2a\ge2$.