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 $f\in C^{\infty}(S^1)$. 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 simple calculation with Riemann sums (using the standard error term in the trapezoid rule) shows $|\hat f(n)-\widehat{f_N}(n)|=O(\frac {n^2}{N^2})$. Since $f$ is smooth, $\hat f(n)=O(\frac 1{n^2})$ and therefore $|\widehat{f_N}(n)|=O(\frac 1{n^2}+\frac {n^2}{N^2})$.
It follows that $\sup_{S^1}|f-f_N|=O(

Since $f$ and $f_N$ coincide on $S_N$ and the gaps in $S_N$ have size $\frac 1N$, we obtain that $\sup_{S^1}|f-f_N|=O(\frac 1n+\frac{n^2}