= Solution
The original PDF's kernel has an incorrect normalization. With ordinary $dt$, its integral is $\pi$, not one. Already at $n=1$ the printed kernel is the constant $1/2$. For $f\equiv1$, the printed operator therefore returns $\pi$, while $\omega(f,\delta)=0$ for every $\delta$. Thus \b[the literal inequality with the printed kernel is false]. The <Fejér kernel> representing the stated average of <Fourier partial sums> is
$$
K_n(t)=\frac1{2\pi n}\left(\frac{\sin(nt/2)}{\sin(t/2)}\right)^2,
\qquad \int_{-\pi}^{\pi}K_n(t)\,dt=1.
$$
The ratio at zero is interpreted by its limit. To verify the normalization and the averaging property, use the finite geometric sum:
$$
K_n(t)=\frac1{2\pi n}\left|\sum_{j=0}^{n-1}e^{ijt}\right|^2
=\frac1{2\pi}\sum_{|r|<n}\left(1-\frac{|r|}{n}\right)e^{irt}.
$$
<Fourier orthogonality> gives unit integral; the displayed <Fourier coefficients> are exactly the multipliers of the <Fejér sum>. This also proves nonnegativity.
For the correctly normalized <Fejér sum>, subtraction of $f(x)$ gives
$$
|\sigma_{n-1}f(x)-f(x)|\le\int_{-\pi}^{\pi}K_n(t)|f(x-t)-f(x)|\,dt.
$$
Take $0<\delta\le1$ and split at $|t|=\delta$. The inner part is at most $\omega(f,\delta)$, because the kernel is nonnegative and has mass one. The strict inequality in the definition of the <modulus of continuity> causes no endpoint issue: <continuity> gives the same bound at distance exactly $\delta$.
On $0<|t|\le\pi$, $\sin(|t|/2)\ge|t|/\pi$, so
$$
K_n(t)\le\frac{\pi}{2nt^2},\qquad\int_{\delta\le|t|\le\pi}K_n(t)\,dt\le\frac{\pi}{n\delta}.
$$
Divide an arc of length $|t|$ into at most $1+|t|/\delta$ shorter arcs. The <triangle inequality> then bounds $|f(x-t)-f(x)|$ by $(1+\pi/\delta)\omega(f,\delta)$. Consequently
$$
\|\sigma_{n-1}f-f\|_\infty\le\left[1+\frac{\pi}{n\delta}\left(1+\frac\pi\delta\right)\right]\omega(f,\delta).
$$
Choosing $\delta=n^{-1/2}$ proves the requested <fractional-scale Fejér approximation bound> for the intended normalization, for example with the absolute constant $1+\pi+\pi^2$:
$$
\boxed{\|\sigma_{n-1}f-f\|_\infty\le(1+\pi+\pi^2)\,\omega(f,n^{-1/2}).}
$$
The same proof applies to complex-valued <continuous functions>, since the estimates use absolute values.
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