Is this a valid proof that sine is continuous at the origin? $$
\text{Let } \left|\sin x - 0\right| < \epsilon. \\
-\epsilon < \sin x < \epsilon \\
\arcsin (-\epsilon) < x < \arcsin (\epsilon) \\
-\arcsin \epsilon < x < \arcsin \epsilon \\
\left|x\right| < \arcsin \epsilon \\
\left|x - 0\right| < \arcsin \epsilon \\
\text{Let } \delta = \arcsin \epsilon. \\
0 < \left|x - 0\right| < \delta \implies \left|\sin x - 0\right| < \epsilon \\
\lim_{x->0} \sin x = 0 \\
\lim_{x->0} \sin x = \sin 0 \\
\sin x \text{ is continuous at the origin}
$$
In particular, is it safe to get from $-\epsilon < \sin x < \epsilon$ to $\arcsin (-\epsilon) < x < \arcsin (\epsilon)$ by applying the inverse function to all sides of the inequality? Can this operation be dangerous for some functions, functions whose inverses don't share a strictly positive or negative relation?

Define $\sin x$ on a small neighborhood $(-\delta, \delta) \subset \mathbb{R}$ by its McLaurin series $\sin x = \sum_{n = 0}^\infty (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$. For $x \in \mathbb{R}$ with $|x|$ sufficiently small, the value of the series can be made arbitrarily small...
