Here's what I have, Let $\epsilon>0$. Then there exists N such that $m,n>N \implies |n^2-m^2|<\epsilon.$ So now we have to find a $N$ that makes the implication true. So, $|n^2-m^2|<\epsilon=n^2<\epsilon+m^2=n<\sqrt{\epsilon+m^2}$. However, this can become arbitrarily large for large $m$. Hence, the sequence $s_n$ is not Cauchy.
I figure this can be written more formally, but I just want to know if what I typed is correct.
On
For $n \neq m$, $|n^{2} - m^{2}| \in \mathbb{Z}$, $|n^{2} - m^{2}| \neq 0$. There isn't an integer between $0$ and $1$, so if you pick $\epsilon = \frac{1}{2}$, then $|m^{2} - n^{2}| > \epsilon$ whenever $m \neq n$.
On
You have to show the sequence is not Cauchy. A sequence ${a_n}$is Cauchy if for every $\epsilon>0$ the statement: "There exists $N$ such that $n,m>N$ $\implies |a_n-a_m| < \epsilon$" is true.
So, to show the sequence is not Cauchy, we need to show that for atleast one particular number $\alpha$, the statement : "There exists $N$ such that $n,m>N$ $\implies |a_n-a_m| < \alpha$" is false. This means for that $\alpha$ : "For all $N$ there exist $n,m>N$ such that $|a_n-a_m| > \alpha$" is true.
So, you have to make a choice of $\alpha$ and prove the statement : "For all $N$ there exist $n,m>N$ such that $|a_n-a_m| > \alpha$"
Assume $\epsilon<1$. Then for any $n\neq m$, $|n^2-m^2|>\epsilon$.
To use your proof, you could also assume that $\epsilon\leq 1$. Then $$n<\sqrt{\epsilon+m^2}\leq \sqrt{(1+m)^2}=1+m$$ is not true if $n\geq m+1$, which also shows that this is not a Cauchy sequence.
It is easiest to pick a value of $\epsilon$ where we can never pick the desired $n$ and $m$. You can also prove it can exceed any $\epsilon$ by choosing $n>m+\epsilon+1$, so that $|n^2-m^2|>m^2+2(\epsilon+1)m+(\epsilon+1)^2-m^2=2(\epsilon+1)m+(\epsilon+1)>\epsilon$.