Here is a solution that I propose for the statement of the title.
Recall two facts:
$$ \displaystyle \cos(a) - \cos(b) = -2\sin\left (\frac{a+b}{2} \right)\sin\left (\frac{a-b}{2} \right) $$ and $$ \forall x\in [0,\infty), \ \sin(x) \le x $$
Now, let $\epsilon >0$ arbitrary and choose $N= \displaystyle \left \lceil \sqrt[4]{\frac{1}{2\epsilon}} \right \rceil \ge \sqrt[4]{\frac{1}{2\epsilon}}$.
If $n > N$ then (after a bit of algebra) you get that $\displaystyle \frac{1}{2n^4} < \epsilon$. Observe now that:
$$ \displaystyle \left | \cos \left ( \frac{1}{n^2}\right ) -1 \right| = \left | \cos \left ( \frac{1}{n^2}\right ) -\cos(0) \right| = \left | 2\sin^2 \left ( \frac{1}{2n^2}\right ) \right| = 2\sin^2 \left ( \frac{1}{2n^2}\right ) \le 2\left (\frac{1}{2n^2} \right )^2 = \frac{1}{2n^4} $$
This concludes the proof. I have not checked each of the steps I am making thoroughly so I would it appreciate it if you can let me know of any inconsistencies. Please feel free to rearrange the proof its self so as to make it optimally legible.