= Solution
Use the centered <second modulus of smoothness>
$$
\omega_2(f,\delta)=\sup_{|h|\le\delta}\|f(\cdot+h)-2f+f(\cdot-h)\|_\infty.
$$
The relevant modulus properties are $\omega_2(f,t)\le(1+t/\delta)^2\omega_2(f,\delta)$ for $t\ge0$, $\delta>0$, and $\omega_2(f,\delta)\le\delta^2\|f''\|_\infty$ when $f''$ is continuous. The latter follows from the <second-difference integral formula>
$$
f(x+h)-2f(x)+f(x-h)=\int_{-|h|}^{|h|}(|h|-|u|)f''(x+u)\,du.
$$
The <Fejér kernel> is even, nonnegative and has integral one. Averaging the convolution at $t$ and $-t$ therefore gives
$$
\sigma_nf(x)-f(x)=\frac12\int_{-\pi}^{\pi}F_n(t)\bigl(f(x+t)-2f(x)+f(x-t)\bigr)\,dt.
$$
Apply the <second modulus of smoothness> bound and $(1+s)^2\le2(1+s^2)$:
$$
\|\sigma_nf-f\|_\infty\le\omega_2(f,\delta)\left(1+\delta^{-2}\int_{-\pi}^{\pi}t^2F_n(t)\,dt\right).
$$
For $|t|\le\pi$, $\sin(|t|/2)\ge|t|/\pi$, while $\sin^2(nt/2)\le1$. Thus
$$
t^2F_n(t)\le\frac{\pi}{2n},\qquad \int_{-\pi}^{\pi}t^2F_n(t)\,dt\le\frac{\pi^2}{n}.
$$
The removable value at $t=0$ causes no problem. With $\delta=n^{-1/2}$, the <Fejér second-modulus approximation bound> is
$$
\boxed{\|\sigma_nf-f\|_\infty\le(1+\pi^2)\omega_2(f,n^{-1/2}).}
$$
If $f''$ is continuous, this also yields \b[$\|\sigma_nf-f\|_\infty\le(1+\pi^2)\|f''\|_\infty/n=O(n^{-1})$].
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