A subset $S$ of $\mathbb{R}$ is complete if every Cauchy sequence consisting of elements of $S$ converges to an element of $S$.
Prove that the following set $A = \{x \in \mathbb{R} \setminus\mathbb{Q}\mid x^2 \leq \frac{1}{4} \} \subset \mathbb{R}$ is complete.
Note: I don't want the solution (at least not eight now) but I want to run past people my idea of how to prove this.
IDEA:
I feel that this set is not complete. I'm basing the idea off of the same notion of how the rationals $\mathbb{Q}$ are not complete. Particularly I could have a sequence converging to $\sqrt{2}$. In this problem I want to use that same approach. Based on the conditions of my set
$$ \frac{-1}{2} \leq x \leq \frac{1}{2}$$
So I was thinking that I could create a sequence that would converge to $\frac{1}{2}$. And by virtue of it converging it means it is Cauchy. To create this sequence I would use the idea that the product of an irrationl and rational number is an irrational number. An idea of what I would like the sequnce terms to look like would be something of this vein:
$$ x_n = \Bigg(\frac{n-1}{n}\Bigg)\Bigg(\frac{1}{\sqrt{2}}\Bigg) $$
I know this specific term doesn't work. What I was hoping to accomplish was since $\lim_{n \rightarrow \infty}\frac{n-1}{n} \rightarrow 1$, I could somehow use that product of terms to converge to $\frac{1}{2}$ What would be ideal was if I could somehow manufacture the irrational terms converge to 1 and have $\frac{1}{2}$ as a constant, but I don't know how I could possibly create a sequence of irrationals without having an explicit irrational value in it.
Am I on the right path? What could I do to help my situation?
If you want an explicit sequence, you could take the sequence $a_n = 1/2 - 1/\sqrt{p_n}$ with $p_n$ the $n$'th prime number.