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This is an exercise in writing out the definitions: since the defect group of $B$ is $E$, we have $B|(B_{\delta(E)})^{G\times G}$. So by assumption, $$ b\;|\;B_{H\times H}\;|\;\left((B_{\delta(E)})^{G\times G}\right)_{H\times H}=\bigoplus_{(g_1,g_2)\in \delta(E)\backslash G\times G/H\times H}\left(B_{\delta(E)^{(g_1,g_2)}\cap H\times H}\right)^{H\times H}, $$ which means that $b$ divides one of the summands (since $b$ is simple). Say $b|M^{H\times H}$, where $M$ is a $(\delta(E)^{(g_1,g_2)}\cap H\times H)$-module. So the vertex of $b$ is some subgroup of that intersection, and in particular a subgroup of $\delta(E)^{(g_1,g_2)}$, as required.