Showing that for all rational $\epsilon > 0$, there is a rational $x\ge 0$ with $x^2 < 2 < (x+\epsilon)^2$ This question is related to Tao's Analysis I.  In the book, there is such a proposition:

Proposition 4.4.5: For every rational number $\epsilon>0$, there exists a non-negative rational number $x$ such that $x^2<2<(x+\epsilon)^2$.

Can I prove it as follows:

Let $\epsilon>0$ be rational. Suppose for the sake of contradiction that there is no non-negative rational number $x$ for which $x^2<2<(x+\epsilon)^2$. This means that whenever $x$ is non-negative and $x^2<2$, we must also have $(x+\epsilon)^2<2$
(note that $(x+\epsilon)^2$ cannot be equal to $2$ )
Take $\epsilon=3$. Clearly $(x+3)^2>2$ for all non-negative rational numbers $x$, a contradiction.

 A: No. In

Let $\epsilon > 0$ be rational

you have already chosen some arbitrary $\epsilon$. You cannot take $\epsilon = 3$.
A: No. A proof by contradiction would proceed as follows:
Suppose the proposition were false. Then there would exist some rational $\epsilon > 0$ such that for no rational $x$ we had that $x^2 < 2 < (x + \epsilon)^2$. And from that you would try to derive a contradiction.
But you do not know what this hypothetical $\epsilon$ would be. What you write only tells us that it cannot be $3$.
This proposition calls for a constructive proof: You are given a rational $\epsilon > 0$ and should then find a rational non-negative $x$ such that $x^2 < 2 < (x + \epsilon)^2$. Hint: Try $x = 2 - \frac{p}{2q}$, if $\epsilon = \frac{p}{q}$.
A: No, your proof is not correct. You need to prove that the result is valid for any $\epsilon$. So you can’t select $\epsilon =3$ like you did.
You don’t need to proceed by contradiction. Just notice  that the largest $n\epsilon$  with $n \in \mathbb N$ such that $(n\epsilon)^2<2$ would satisfy the required condition.
