= Solution
Use $\mathbb T=\mathbb R/(2\pi\mathbb Z)$ and normalized <Haar measure> $dm=ds\,dt/(2\pi)^2$ on the <torus>. Here $|s|$ means distance to zero modulo $2\pi$, represented in $[-\pi,\pi]$; an ordinary coordinate in $[0,2\pi)$ would not express concentration at the origin correctly.
Put $B(s,t)=2+\cos s+\cos t$, $I_N=\int B^Ndm$, and $A_N=I_N^{-1}$. Since $0\le B\le4$ and $B(0,0)=4$, $I_N>0$, the kernel $K_N=A_NB^N$ is nonnegative, and $\int K_Ndm=1$ exactly. Fix $0<\delta\le\pi$. On the set where either $|s|\ge\delta$ or $|t|\ge\delta$, $B\le q=3+\cos\delta<4$. Choose $q'$ with $q<q'<4$. <continuity> provides a neighborhood $U$ of the origin of positive measure $v$ where $B\ge q'$. Consequently
$$
I_N\ge v(q')^N,\qquad\sup_{\max(|s|,|t|)\ge\delta}K_N(s,t)\le\frac1v\left(\frac q{q'}\right)^N\longrightarrow0.
$$
For $0<\varepsilon\le\pi$, take $\delta=\varepsilon$ and then $N$ large enough that this bound is at most $\varepsilon$. For $\varepsilon>\pi$ the exterior conditions are empty. This proves all the requested kernel properties, with $\boxed{A=\left[\int(2+\cos s+\cos t)^Ndm\right]^{-1}}$.
Now let $M=\|f\|_\infty$ and define $P_N=K_N*f$. The <powered-cosine approximate identity on a torus> is a <trigonometric polynomial>, since its base is a finite sum of <complex exponentials> and its exponent is an integer. Expanding $K_N$ and integrating its finitely many Fourier terms shows that $P_N$ is also a <trigonometric polynomial>; it is real-valued because both factors are real. Translation invariance and normalization give
$$
P_N(x,y)-f(x,y)=\int K_N(s,t)[f(x-s,y-t)-f(x,y)]dm(s,t).
$$
Given $\eta>0$, <uniform continuity> of $f$ supplies $\delta>0$ making the difference in brackets smaller than $\eta/2$ on the square $|s|,|t|<\delta$, uniformly in $(x,y)$. The contribution from that square is at most $\eta/2$. On its complement it is at most $2M\sup_{\max(|s|,|t|)\ge\delta}K_N$, which tends to zero by the preceding estimate. Therefore $\boxed{\|P_N-f\|_\infty\to0}$, proving uniform approximation without assuming any Fourier-series convergence of $f$.
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