Unfortunately, there seems to be a counterexample: Let $B^p_i$ be the interval $(p-i,\infty)$ in the integers $\mathbb{Z}$, for natural numbers $p$ and $i$. In this case, we have $B^{p+1}_i\subset B^p_i$, and for any fixed $i$, we have $\bigcap_p B^p_i=0$, since $p$ runs out to infinty. But for fixed $p$, the limit of $B^p_i$ as $i$ increases is all of $\mathbb{Z}$. Thus, the intersection of these limits is also $\mathbb{Z}$, which is not empty.
A simpler counterexample, using the same idea: let $B^p_i=\{1\}$B^p_i=A$ for some fixed non-empty$A$, if$p\lt i$, otherwise$B^p_i=$emptyset. In this case, for fixed$i$we have$\bigcap_p B^p_i=0$, since eventually$p$exceeds$i$, but for fixed$p$, the$B^p_i$are eventually equal to {1}$A$as$i$increases, and so have limit {1}.$A$. 2 added 303 characters in body Unfortunately, there seems to be a counterexample: Let$B^p_i$be the interval$(p-i,\infty)$in the integers$\mathbb{Z}$, for natural numbers$p$and$i$. In this case, we have$B^{p+1}_i\subset B^p_i$, and for any fixed$i$, we have$\bigcap_p B^p_i=0$, since$p$runs out to infinty. But for fixed$p$, the limit of$B^p_i$as$i$increases is all of$\mathbb{Z}$. Thus, the intersection of these limits is also$\mathbb{Z}$, which is not empty. A simpler counterexample, using the same idea: let $B^p_i=\{1\}$ if$p\lt i$, otherwise$B^p_i=$emptyset. In this case, for fixed$i$we have$\bigcap_p B^p_i=0$, since eventually$p$exceeds$i$, but for fixed$p$, the$B^p_i$are eventually equal to {1} as$i$increases, and so have limit {1}. 1 Unfortunately, there seems to be a counterexample: Let$B^p_i$be the interval$(p-i,\infty)$in the integers$\mathbb{Z}$, for natural numbers$p$and$i$. In this case, we have$B^{p+1}_i\subset B^p_i$, and for any fixed$i$, we have$\bigcap_p B^p_i=0$, since$p$runs out to infinty. But for fixed$p$, the limit of$B^p_i$as$i$increases is all of$\mathbb{Z}$. Thus, the intersection of these limits is also$\mathbb{Z}\$, which is not empty.