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?

Provided you know properties of the arcsine your idea will be a proof.  However, are you sure you do not need to know that $\sin$ is continuous to deduce properties of the arcsine?  
Spivak's calculus book has a note about a faulty proof he had in there in one of the pre-publication drafts.  It used the square-root function in a proof that $x^2$ is continuous.  But then, later, he used continuity of $x^2$ in the proof that the square-root exists.  Fortunately, he caught the mistake before publication.
