B

连接$BE$、$CD$交于点$F$，$CD$交$AB$于点$H$，则$\angle FHB=\angle AHC$，
$\because \angle BAC=\angle DAE=90^{\circ}$，$AB=AC=2$，$AD=AE=1$，
$\therefore \angle BAE=\angle CAD=90^{\circ}+\angle BAD$，$BC^{2}=AB^{2}+AC^{2}=2^{2}+2^{2}=8$，$DE^{2}=AD^{2}+AE^{2}=1^{2}+1^{2}=2$，
在$\triangle BAE$和$\triangle CAD$中，
$\left\{\begin{array}{l}{AB=AC}\\{∠BAE=∠CAD}\\{AE=AD}\end{array}\right.$，
$\therefore \triangle BAE$≌$\triangle CAD\left(SAS\right)$，
$\therefore \angle ABE=\angle ACD$，
$\therefore \angle BFD=\angle ABE+\angle FHB=\angle ACD+\angle AHC=90^{\circ}$，
$\therefore \angle CFE=\angle BFC=\angle DFE=90^{\circ}$，
$\therefore BD^{2}=BF^{2}+DF^{2}$，$CE^{2}=CF^{2}+EF^{2}$，$BC^{2}=BF^{2}+CF^{2}$，$DE^{2}=DF^{2}+EF^{2}$，
$\therefore BD^{2}+CE^{2}=BC^{2}+DE^{2}=BF^{2}+DF^{2}+CF^{2}+EF^{2}$，
$\because BC^{2}+DE^{2}=8+2=10$，
$\therefore BD^{2}+CE^{2}=10$，
故选：$B$.