I had to revise the answer somewhat :-). Probably it is in fact not that easy to use the Stone Weierstrass Theorem for this, but the approach using Fourier Series works out fine.
The $N$-th Fourier polynomial of a continuous periodic function (on $[-\pi,\pi]$, say) is given by $ S_N(f)(x) := \sum_{k=0}^N c_k(f) e^{ikx}$ where $ c_k(f) := \frac{1}{2\pi} \int_{-\pi}^{\pi}f(y) e^{-iky}dy$
The Fourier Series of $f$ corresponds to $N=\infty.$ If $f$ is differentiable and f^' integrable such the fundamental theorem of calculus holds, it is easy to see (by partial integration) that \frac{d}{dx}S_N(f)(x)= \sum_{k=0}^Nc_k(f') e^{-ikx} that is, the derivative of the $N$-th Fourier polynomial is the $N$-th Fourier polynomial of the derivative. (All this can be found in Rudin's "Principles of Mathematical Analysis").
While, in general, it will not be true that the Fourier series of a $C^k$ function $f$ converges together with the first $k$ derivatives uniformly to $f$ and it's derivatives, this will be true if $f$ is, e.g., piecewise $C^{k+1}$. This is shown in most textbooks on analysis for piecewise differentiable $f$ (e.g. in Rudin's book mentioned above), and because of the aforementioned relation between fourier coefficients and taking the derivatives this result may be applied to $f^{(k)}$ to prove the same statement for piecewise $C^{k+1}$ functions.
But piecewise $C^{k+1}$ functions are dense in $C^k$. Consequently, trigonometric polynomials are, too.