Prove $\tan( \frac {\pi x}{2})$ from $(-1,1)$ to $\mathbb R$ is continuous. Prove $\tan( \frac {\pi x}{2})$  from $(-1,1)$ to $\mathbb R$ is continuous using the open  sets definition.
Proof. So Let $S$ an open set in $R$ so $S=\cup_{i\in I}(a,b)$ .I wanna prove that $$f^{-1}(S)={(x \in (-1,1)| f(x) \in S )}$$ is open in $(-1,1)$.
Let $x$ such that $f(x) \in S$ .Then there exist an open set in $\mathbb R$ such that $$f(x) \in (f(x)-δ_χ,f(x)+δ_x)$$ so $$\tan( \frac {\pi x}{2}) -δ_x < \tan( \frac {\pi x}{2}) < \tan( \frac {\pi x}{2}) +δ_x$$
Now my thought was to solve for x so I can find and $$f^{-1}((f(x)-δ_χ,f(x)+δ_x))$$ for every $x$ and say it is open in $(-1,1)$  subset of $f^{-1}(S)$ so since for every $x \in f^{-1}(S) $.  I have an open area of it, but I can't solve the inequality.
 A: You can simplify the problem showing that $f = \tan $ is continuous in $(-\pi/2,\pi/2)$. 
Let us fix an interval $(a,b) \subset \mathbb{R}$. You want to show that $\tan^{-1}((a,b))$ is open in $(-\pi/2,\pi/2)$. Since $\tan$ is a bijective function in that interval (it's strictly increasing), then $\tan^{-1}((a,b)) = (\arctan a, \arctan b)$, which is open. Since $f^{-1}(\bigcup_i A_i) = \bigcup_i f^{-1}(A_i)$ (check!), then pre-images of any open set (arbitrary union of intervals) are still open. Hence $\tan $ is continuous. 
A very similar technique can be applied to your case. Now I would conclude saying that $g$ such that $g(x) = \frac{\pi x}{2}, x \in (-1,1)$ is continuous (repeat a similar argument) and then $f \circ g$ (your function) is continuous, because composition of two continuous maps (check!). 
A: All this is bullshit. Either you believe (or are allowed to use) that $\tan$ is continuous on $\left]{-{\pi\over2}},{\pi\over2}\right[\>$ or not. If yes, then $\tan\circ\bigl({\pi\over2}{\rm id}\bigr)$ is continuous as well. If no, then go back and prove that $\sin$, $\cos$, and the quotient of two continuous functions (as long as the denominator is $\ne0$) are continuous.
