To prove that one set is a subset of another, we start with an element of the smaller set and see if we can determine that it is a member of the larger set (Definition SSET). Suppose $\vect{x}\in\nsp{B}$. Then we know that $B\vect{x}=\zerovector$ by Definition NSM. Consider
\begin{align*}
(AB)\vect{x}&=A(B\vect{x})&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\
&=A\zerovector&&\text{Hypothesis}\\
&=\zerovector&&\knowl{./knowls/theorem.MMZM.knowl}{\text{Theorem MMZM}}\\
\end{align*}
This establishes that $\vect{x}\in\nsp{AB}$, so $\nsp{B}\subseteq\nsp{AB}$.
To show that the inclusion does not hold in the opposite direction, choose $B$ to be any nonsingular matrix of size $n$. Then $\nsp{B}=\set{\zerovector}$ by Theorem NMTNS. Let $A$ be the square zero matrix, $\zeromatrix$, of the same size. Then $AB=\zeromatrix B=\zeromatrix$ by Theorem MMZM and therefore $\nsp{AB}=\complex{n}$, and is not a subset of $\nsp{B}=\set{\zerovector}$.