= Solution
Use real physical fields in the quadratic energy expressions. The vector identity $\nabla\cdot(\mathbf E\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf B)$ and vacuum <Maxwell's equations> give
$$
\nabla\cdot\left(\frac{\mathbf E\times\mathbf B}{\mu_0}\right)
=-\frac1{\mu_0}\mathbf B\cdot\partial_t\mathbf B-\varepsilon_0\mathbf E\cdot\partial_t\mathbf E
=-\partial_t\left(\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}\right).
$$
Integrating over a fixed volume and applying the <divergence theorem> proves <Poynting's theorem>:
$$
\boxed{\oint_A\frac{\mathbf E\times\mathbf B}{\mu_0}\cdot d\mathbf A
=-\frac d{dt}\int_V\left(\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}\right)dV.}
$$
The left side is outward energy flux through the closed surface, not merely local intensity.
For a linearly polarized wave with phase $\chi=\mathbf k\cdot\mathbf r-\omega t$, the real fields give <Poynting vector> $\mathbf S=\varepsilon_0cE_0^2\cos^2\chi\,\widehat{\mathbf k}$. Local intensity oscillates at $2\omega$. In the closed-surface integral, the time-independent part cancels because $\oint_A\widehat{\mathbf k}\cdot d\mathbf A=0$. The remaining flux generally oscillates at $2\omega$, has zero time average, and may vanish identically for special volume geometries.
For <circular polarization>, take $\mathbf E=E_0(\mathbf e_1\cos\chi+\mathbf e_2\sin\chi)$ with orthonormal transverse vectors. Then $E^2=E_0^2$ and $\mathbf B=\widehat{\mathbf k}\times\mathbf E/c$, so $\mathbf S=\varepsilon_0cE_0^2\widehat{\mathbf k}$ is spatially and temporally constant. Therefore the <closed-surface Poynting flux of a plane wave> is
$$
\boxed{\oint_A\mathbf S\cdot d\mathbf A=0\quad\text{at every time for circular polarization}.}
$$
There is still a nonzero steady local energy flow: equal energy enters and leaves every fixed closed volume. Calling the net closed-surface flux a nonzero constant would confuse these two quantities.
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