Here is a proof sketch along the lines suggested in the linked SE answer. 

I will work on $S^1=[0,1]$ (identifying $0$ and $1$) and denote $e(x)=e^{2\pi ix}$. I will also denote $S_N=\{0,\frac 1N,\ldots,\frac{N-1}N\}\subset S^1$ and assume for convenience that $N$ is odd. Let $f\in C^{2}(S^1)$. Set $f_N(x)=\sum_{n=-(N-1)/2}^{(N-1)/2}\widehat {f_N}(n)e(nx)$, where $\widehat{f_N}(n)=\frac 1N\sum_{x\in S_N}f(x)e(-nx)$. Then $f_N$ coincides with $f$ on $S_N$ by the Discrete Fourier Transform. 

A calculation using Abel's summation formula twice and that $f\in C^2$ shows $\widehat{f_N}(n)=O(\frac 1{n^2})$ for $-\frac{N-1}2\le n\le\frac{N-1}2$, uniformly in $N$. This is the discrete version of the (hopefully "intuitive") fact that the Fourier coefficients of a smooth function decay fast. The actual calculation with this approach involves expressions quite similar to Dirichlet and Fejer kernels, evaluated at a discrete set of points. An alternative, more elegant approach is to use the identity $\widehat{\Delta g}(n)=(e(\frac nN)-1)\hat g(n)$ for discrete Fourier transforms, where $\Delta$ is the difference operator $\Delta g(\frac kN)=g(\frac{k+1}N)-g(\frac kN)$.

From this one obtains $\sup_{S^1}|f_N'|=O(\log N)$ and since $f=f_N$ on $S_N$ and the gaps between the points of $S_N$ are $1/N$ we have that $\sup_{S^1}|f-f_N|=O(\log N/N)$ and we are done.