Umm...so what?
I am so lost on this proof. I know $B^+$ is $BB^*$ or $B^+$ has all the strings that are 1 or more concatenations of strings from $B$. This is what I managed to logic so far.
if $B = B^+$, then $BB \subseteq B^+$.
I have no clue where to go from here. Can anyone point me in the right direction?
The condition $B^2 \subseteq B$ says that $B$ is closed under (concatenation) product and $B^+$ denotes the semigroup generated by $B$. Thus both conditions simply mean that $B$ is a subsemigroup of the free monoid and hence are equivalent.