Sequence Convergence of $(s_n) = (\frac{1}{n^2})$ Part 1. Given $\varepsilon = 0.5$, find an $N \in \mathbb{N}$ s.t whenever $n \geq N$, $|s_n - 0| < \varepsilon$
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Part 2. Treat the epsilon in Part 1 with an arbitrary $\varepsilon > 0$

My work:
Let $s_n = \frac{1}{n^2}$, and suppose $s = 0$, implies $|\frac{1}{n^2} - 0| < \varepsilon$. Now is we assume $\varepsilon = 0.5$, then:
$|\frac{1}{n^2} - 0| < 0.2 \Rightarrow \frac{1}{n^2} < 0.5 \Rightarrow n^2 > \frac{1}{0.2} \Rightarrow n > \frac{1}{\sqrt{0.5}}$, we'll chose a $N$ s.t. $N > \frac{1}{\sqrt{0.5}}$.
So now by the Archimedes Prop of $\mathbb{R}$ let $N \in \mathbb{N}$ satisfying $N > \frac{1}{\sqrt{0.5}}$. 
Then if 
$n \geq N, |s_n - s| = |\frac{1}{n^2} - 0| = \frac{1}{n^2} \leq \frac{1}{N^2}$
$\frac{1}{(\frac{1}{\sqrt{0.5}})^2} = \frac{1}{\frac{1}{0.5}} = 0.5 = \varepsilon$
Part 2. Let $\varepsilon > 0$. By the Arch. Prop. of $\mathbb{R}$, let $N \in \mathbb{N}$ s.t. $N > \frac{1}{\sqrt{\varepsilon}}$. Now if $n \geq N$, then:
$|s_n - s| = |\frac{1}{n^2} - 0|$
$= \frac{1}{n^2}$ (since $n > 0$)
$\leq \frac{1}{N^2}$ (since $n \geq N)$
$< \frac{1}{(\frac{1}{\sqrt{\varepsilon}})^2}$ (since $N > \frac{1}{\sqrt{\varepsilon}})$
$= \varepsilon$

My issue is the fact that $N \in \mathbb{N}$, as that throws a problem in my part 1, where did go wrong?

Just use the ceiling function, we just have to choose an $N \in \mathbb{N}$ such that $N> \frac{1}{\sqrt{\epsilon}}$, you can choose it to be $1+\lceil \frac{1}{\sqrt{\epsilon}}\rceil.$
