Since \begin{align} \operatorname{Cov}(X_i-\bar X,\bar Y)&=\operatorname{Cov}(X_i,\bar Y)-\operatorname{Cov}(\bar X,\bar Y) \\&=\frac1n\rho -\frac1{n^2}n\rho \\&=0 \end{align} the vectors $$ (X_1-\bar X,X_2-\bar X,\dots,X_n-\bar X) \tag{1} $$ and $$ \bar Y \tag{2} $$ are independent since they are jointly multivariate normal. Hence, $\sum_{i=1}^n (X_i-\bar X)^2$ (a function only of (1)) is also independent of (2).

Since \begin{align} \operatorname{Cov}(X_i-\bar X,\bar Y)&=\operatorname{Cov}(X_i,\bar Y)-\operatorname{Cov}(\bar X,\bar Y) \\&=\frac1n\rho -\frac1{n^2}n\rho \\&=0 \end{align} the vectors $$ (X_1-\bar X,X_2-\bar X,\dots,X_n-\bar X) \tag{1} $$ and $$ \bar Y \tag{2} $$ are independent since they are jointly multivariate normal. Hence, $\sum_{i=1}^n (X_i-\bar X)^2$ (a function of (1)) is also independent of (2).