Expand $\|\delta^{JS}(X) - \theta\|^2 = \|X - \theta - (p-2)\|X\|^{-2} X\|^2$ by writing $\delta^{JS}(X) - \theta = (X - \theta) - (p-2)\|X\|^{-2}X$. Taking expectation and expanding the square yields three terms: \begin{align*} R(\delta^{JS}, \theta) &= \mathbb{E}_\theta[\|X - \theta\|^2] + (p-2)^2 \, \mathbb{E}_\theta\!\left[\frac{\|X\|^2}{\|X\|^4}\right] - 2(p-2)\,\mathbb{E}_\theta\!\left[\frac{X^\top(X - \theta)}{\|X\|^2}\right] \\ &= p + (p-2)^2 \, \mathbb{E}_\theta\!\left[\frac{1}{\|X\|^2}\right] - 2(p-2)\,\mathbb{E}_\theta\!\left[\frac{X^\top(X - \theta)}{\|X\|^2}\right]. \end{align*} The key step is to evaluate the cross term $\mathbb{E}_\theta[X^\top(X-\theta)/\|X\|^2]$ by applying Stein's lemma coordinate by coordinate. Write this as a sum over $j = 1, \ldots, p$, condition on $X_{(-j)} = (X_1, \ldots, X_{j-1}, X_{j+1}, \ldots, X_p)$ via the tower property, and define $g_j(x) = x/(x^2 + \sum_{i \neq j} X_i^2)$. Since $X_j \sim N(\theta_j, 1)$ given $X_{(-j)}$ and $g_j$ is bounded with bounded derivative (the derivative can be computed explicitly and is bounded a.s.), Stein's lemma gives $\mathbb{E}[(X_j - \theta_j)g_j(X_j)] = \mathbb{E}[g_j'(X_j)]$. Computing $g_j'$ and summing over $j$ gives $\mathbb{E}_\theta[X^\top(X-\theta)/\|X\|^2] = (p-2)\,\mathbb{E}_\theta[1/\|X\|^2]$. Substituting back yields the formula $R(\delta^{JS}, \theta) = p - (p-2)^2 \,\mathbb{E}_\theta[1/\|X\|^2]$. The term $\mathbb{E}_\theta[1/\|X\|^2]$ is strictly positive because the integrand is positive and has positive mass on any annulus $\{c_1 \leq \|x\| \leq c_2\}$, completing the proof.