The proof applies the tower property of conditional expectation in two directions, then combines the resulting equalities. Recall that for any function $Z(X, \theta)$ under the joint law $Q$, the tower property gives both \begin{align*} \mathbb{E}_Q[Z(X, \theta)] = \mathbb{E}_Q[\mathbb{E}_\Pi[Z(X, \theta) \mid X]] \quad \text{and} \quad \mathbb{E}_Q[Z(X, \theta)] = \mathbb{E}_Q[\mathbb{E}_\theta[Z(X, \theta)]], \end{align*} where the first conditions on $X$ and averages over $\theta$, and the second conditions on $\theta$ and averages over $X$. Set $Z(X, \theta) = \theta\, \delta(X)$. From the first form of the tower property, since $\delta$ is Bayes for quadratic loss it satisfies $\delta(X) = \mathbb{E}_\Pi[\theta \mid X]$, so \begin{align*} \mathbb{E}_Q[\theta\, \delta(X)] = \mathbb{E}_Q[\mathbb{E}_\Pi[\theta\, \delta(X) \mid X]] = \mathbb{E}_Q[\delta(X)\, \mathbb{E}_\Pi[\theta \mid X]] = \mathbb{E}_Q[\delta(X)^2]. \end{align*} From the second form, using unbiasedness $\mathbb{E}_\theta[\delta(X)] = \theta$: \begin{align*} \mathbb{E}_Q[\theta\, \delta(X)] = \mathbb{E}_Q[\mathbb{E}_\theta[\theta\, \delta(X)]] = \mathbb{E}_Q[\theta\, \mathbb{E}_\theta[\delta(X)]] = \mathbb{E}_Q[\theta^2]. \end{align*} Therefore $\mathbb{E}_Q[\delta(X)^2] = \mathbb{E}_Q[\theta^2]$, and expanding: \begin{align*} \mathbb{E}_Q[(\delta(X) - \theta)^2] = \mathbb{E}_Q[\delta(X)^2] - 2\mathbb{E}_Q[\theta\, \delta(X)] + \mathbb{E}_Q[\theta^2] = \mathbb{E}_Q[\theta^2] - 2\mathbb{E}_Q[\theta^2] + \mathbb{E}_Q[\theta^2] = 0. \end{align*} Since the integrand $(\delta(X) - \theta)^2$ is non-negative and integrates to zero under $Q$, it must equal zero $Q$-almost everywhere, meaning $\delta(X) = \theta$ with $Q$-probability one.