"Are there any other well-known periodic functions?"
In one sense, the answer is "no". Every reasonable periodic complex-valued function $f$ of a real variable can be represented as an infinite linear combination of sines and cosines with periods equal the period $\tau$ of $f$, or equal to $\tau/2$ or to $\tau/3$, etc. See Fourier series.
There are also doubly periodic functions of a complex variable, called elliptic functions. If one restricts one of these to the real axis, one can find a Fourier series, but one doesn't do such restrictions, as far as I know, in studying these functions. See Weierstrass's elliptic functions and Jacobi elliptic functions.
Consider $f:\mathbb{R^2}\to\mathbb{R}$
$$f(x_0+h,y_0+k)-f(x_0,y_0)=\\=f(x_0+h,y_0+k)-f(x_0+h,y_0)+f(x_0+h,y_0)-f(x_0,y_0)=$$
$$\text{by MVT} \quad\exists c,d\in\mathbb{R} \quad \text{such that}$$
$$=kf_y(x_0+h,y_0+c)+hf_x(x_0+d,y_0)=$$
$$=kf_y(x_0,y_0)+k[f_y(x_0+h,y_0+c)-f_y(x_0,y_0)]+\\+hf_x(x_0,y_0)+h[f_x(x_0+d,y_0)-f_x(x_0,y_0)]=$$
$$=hf_x(x_0,y_0)+kf_y(x_0,y_0)+r(h,k)$$
with:
$$r(h,k)=h[f_x(x_0+d,y_0)-f_x(x_0,y_0)]+k[f_y(x_0+h,y_0+c)-f_y(x_0,y_0)]$$
and
$$\lim_{(h,k)\to(0,0)} \frac{r(h,k)}{\sqrt {h^2+k^2}}=0 \quad \square$$
The same proof can easily extend to $f:\mathbb{R^n}\to\mathbb{R}$ by adding and subtracting n-1 terms in order to apply MVT $n$ times.
The extension to $f:\mathbb{R^n}\to\mathbb{R^m}$ is immediate by the same proof for each component.
Best Answer
A more formal way to show this is by induction. We know that $f(x) = \sin(x)$ is continuous. Also, $f'(x) = \cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}\cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-\cos(x)) = (-1)^{(n+1)+1}\cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $\frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-\sin(x)) = (-1)^{n+2}\sin(x)$, which is also continuous. So, for all $n \geq 0$, $f^{(n)}(x)$ is continuous.