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?

You can use inequality of:

$$0\le \left| \sin { \left( x \right)  }  \right| \le \left| x \right| $$ 

then appling "squeeze theorem" directly
