How to prove a function is always continuous? For example, how could I prove that $\sin x$ is continuous for ]$-\infty,\infty$[ ?
 A: Let $x \in \mathbb{R}$.
Take $$\sin(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}.$$
Then each of the partial sums are continuous. See here for a proof that the power series expansion converges uniformly (on the interval of convergence.) A sequence of continuous converges uniformly to $\sin(x)$, then $\sin(x)$ is continuous at $x$.
A: Holds more than that, $\sin{x}$ is uniformly continuous by $\varepsilon-\delta$ definition. 
It is enough to choose $\delta=\varepsilon$ and the implication from the definition holds, since
$$|\sin{x}-\sin{a}|=|2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}|<2|\frac{x-a}{2}||1|=|x-a|,$$ where we used known inequalities $\sin{t}<t$ and $\cos{t}<1$.
A: That depends on your definition of sin(x). Is it only intuitive or series or Euler's formula?
For the latest is the proof imho the simplest, using the arithmetic of continuous functions and the fact that the exponential function is a continuous function. 
The proof can be worked out even for the definition via the power series, see the results concerning analytic functions.
A: You could show that $\sin(x)$ belongs to $C^{\infty}(\mathbb{R})$. Indeed the function is infinitely times differentiable. Since a function is which is differentiable must be continuous, the original function is continuous. This proof idea assumes you can take the fact that the derivative of $sin$ is $cos$ for granted. Should you not have that at your disposal, using the power series expansion argument already answered here would be sufficient.
