The argument is very simple. The spacetime metric is
$$
ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu + dz^2 , \qquad z \sim z + w .
$$
We now consider $\langle T_{\mu\nu} \rangle$, $\langle T_{\mu z} \rangle$ and $\langle T_{zz} \rangle$. Each of these quantities is invariant under all the symmetries of the problem and in particular under translations and $d-1$ dimensional Lorentz transformations which ($x^\mu \to \Lambda^\mu{}_\nu x^\nu$). Translational invariance implies that the one-point functions do not depend on coordinates. Further, there is only one Lorentz invariant rank 2 tensor, namely $\eta_{\mu\nu}$ and there is no Lorentz invariant vector. It follows that
$$
\langle T_{\mu\nu} \rangle = A \eta_{\mu\nu} , \qquad \langle T_{\mu z} \rangle = 0 , \qquad \langle T_{zz} \rangle = B .
$$
The dimension of the stress tensor is $d$ so we must have $[A]=[B]=d$. Since the only scale in the problem is $w$, we must have $A=A'/w^d$ and $B=B'/w^d$ where $A'$ and $B'$ are dimensionless. Finally, the stress tensor is traceless so
$$
\eta^{\mu\nu} \langle T_{\mu\nu} \rangle + \langle T_{zz} \rangle = 0 \implies (d-1) A + B = 0 \implies B' = - (d-1)A'
$$
Relabelling $A'\to F$, we have
$$
\langle T_{\mu\nu} \rangle = \frac{F}{w^d} \eta_{\mu\nu} , \qquad \langle T_{\mu z} \rangle = 0 , \qquad \langle T_{zz} \rangle = - ( d -1 ) \frac{F}{w^d} .
$$
