You tried to show detailed balance for the Markov chain that is obtained by considering one transition of the Markov chain to be the 'Gibbs sweep' where you sample each component in turn from its conditional distribution. For this chain, detailed balance is not satisfied. The point is rather that each sampling of a particular component from its conditional distribution is a transition that satisfies detailed balance. It would be more accurate to say that Gibbs sampling is a special case of a slightly generalized Metropolis-Hastings, where you alternate between multiple different proposals. More details follow.
The sweeps do not satisfy detailed balance
I construct a counterexample. Consider two Bernoulli variables ($X_1,X_2$), with probabilities as shown in the following table:
\begin{equation}
\begin{array}{ccc}
& X_2 = 0 & X_2 = 1 \\
X_1 = 0 & \frac{1}{3} & \frac{1}{3} \\
X_1 = 1 & 0 & \frac{1}{3}
\end{array}
\end{equation}
Assume the Gibbs sweep is ordered so that $X_1$ is sampled first. Moving from state $(0,0)$ to state $(1,1)$ in one move is impossible, since it would require going from $(0,0)$ to $(1,0)$. However, moving from $(1,1)$ to $(0,0)$ has positive probability, namely $\frac{1}{4}$. Hence we conclude that detailed balance is not satisfied.
However, this chain still has a stationary distribution that is the correct one. Detailed balance is a sufficient, but not necessary, condition for converging to the target distribution.
The component-wise moves satisfy detailed balance
Consider a two-variate state where we sample the first variable from its conditional distribution. A move between $(x_1,x_2)$ and $(y_1,y_2)$ has zero probability in both directions if $x_2 \neq y_2$ and thus for these cases detailed balance clearly holds. Next, consider $x_2 = y_2$:
\begin{equation}
\pi(x_1,x_2) \mathrm{Prob}((x_1,x_2) \rightarrow (y_1,x_2)) = \pi(x_1,x_2)\,p(y_1 \mid X_2 = x_2) = \pi(x_1,x_2) \, \frac{\pi(y_1,x_2)}{\sum_z \pi(z,x_2)} \\
= \pi(y_1,x_2) \, \frac{\pi(x_1,x_2)}{\sum_z \pi(z,x_2)} = \pi(y_1,x_2) \,p(x_1 \mid X_2 = x_2) = \pi(y_1,x_2) \mathrm{Prob}((y_1,x_2) \rightarrow (x_1,x_2)).
\end{equation}
How the component-wise moves are Metropolis-Hastings moves?
Sampling from the first component, our proposal distribution is the conditional distribution. (For all other components, we propose the current values with probability $1$). Considering a move from $(x_1, x_2)$ to $(y_1, y_2)$, the ratio of target probabilities is
\begin{equation}
\frac{\pi(y_1,x_2)}{\pi(x_1,x_2)}.
\end{equation}
But the ratio of proposal probabilities is
\begin{equation}
\frac{\mathrm{Prob}((y_1,x_2) \rightarrow (x_1,x_2))}{\mathrm{Prob}((x_1,x_2) \rightarrow (y_1,x_2))} = \frac{\frac{\pi(x_1,x_2)}{\sum_z \pi(z,x_2)}}{\frac{\pi(y_1,x_2)}{\sum_z \pi(z,x_2)}} = \frac{\pi(x_1,x_2)}{\pi(y_1,x_2)}.
\end{equation}
So, the ratio of target probabilities and the ratio of proposal probabilities are reciprocals, and thus the acceptance probability will be $1$. In this sense, each of the moves in the Gibbs sampler are special cases of Metropolis-Hastings moves. However, the overall algorithm viewed in this light is a slight generalization of the typically presented Metropolis-Hastings algorithm in that you have alternate between different proposal distributions (one for each component of the target variable).
Best Answer
I am not very familiar with stochastic volatility models, but I do know that in most settings, the reason we choose Gibbs or M-H algorithms to draw from the posterior, is because we don't know the posterior. Often we want to estimate the posterior mean, and since we don't know the posterior mean, we draw samples from the posterior and estimate it using the sample mean. So, I am not sure how you will be able to take the mean from the posterior distribution.
Instead the Rao-Blackwellized estimator depends o the knowledge of the mean of the full conditional; but even then sampling is still required. I explain more below.
Suppose the posterior distribution is defined on two variables, $\theta = (\mu, \phi$), such that you want to estimate the posterior mean: $E[\theta \mid \text{data}]$. Now, if a Gibbs sampler was available you could run that or run a M-H algorithm to sample from the posterior.
If you can run a Gibbs sampler, then you know $f(\phi \mid \mu, data)$ in closed form and you know the mean of this distribution. Let that mean be $\phi^*$. Note that $\phi^*$ is a function of $\mu$ and the data.
This also means that you can integrate out $\phi$ from the posterior, so the marginal posterior of $\mu$ is $f(\mu \mid data)$ (this is not known completely, but known upto a constant). You now want to now run a Markov chain such that $f(\mu \mid data)$ is the invariant distribution, and you obtain samples from this marginal posterior. The question is
How can you now estimate the posterior mean of $\phi$ using only these samples from the marginal posterior of $\mu$?
This is done via Rao-Blackwellization.
\begin{align*} E[\phi \mid data]& = \int \phi \; f(\mu, \phi \mid data) d\mu \, d\phi\\ & = \int \phi \; f(\phi \mid \mu, data) f(\mu \mid data) d\mu \, d\phi\\ & = \int \phi^* f(\mu \mid data) d\mu. \end{align*}
Thus suppose we have obtained samples $X_1, X_2, \dots X_N$ from the marginal posterior of $\mu$. Then $$ \hat{\phi} = \dfrac{1}{N} \sum_{i=1}^{N} \phi^*(X_i), $$
is called the Rao-Blackwellized estimator for $\phi$. The same can be done by simulating from the joint marginals as well.
Example (Purely for demonstration).
Suppose you have a joint unknown posterior for $\theta = (\mu, \phi)$ from which you want to sample. Your data is some $y$, and you have the following full conditionals $$\mu \mid \phi, y \sim N(\phi^2 + 2y, y^2) $$ $$\phi \mid \mu, y \sim Gamma(2\mu + y, y + 1) $$
You run the Gibbs sampler using these conditionals, and obtained samples from the joint posterior $f(\mu, \phi \mid y)$. Let these samples be $(\mu_1, \phi_1), (\mu_2, \phi_2), \dots, (\mu_N, \phi_N)$. You can find the sample mean of the $\phi$s, and that would be the usual Monte Carlo estimator for the posterior mean for $\phi$..
Or, note that by the properties of the Gamma distribution $$E[\phi | \mu, y] = \dfrac{2 \mu + y}{y + 1} = \phi^*.$$
Here $y$ is the data given to you and is thus known. The Rao Blackwellized estimator would then be
$$\hat{\phi} = \dfrac{1}{N} \sum_{i=1}^{N} \dfrac{2 \mu_i + y}{y + 1}. $$
Notice how the estimator for the posterior mean of $\phi$ does not even use the $\phi$ samples, and only uses the $\mu$ samples. In any case, as you can see you are still using the samples you obtained from a Markov chain. This is not a deterministic process.