We will examine the three requirements of Theorem TSS. Recall that Definition NSM can be formulated as $\nsp{A}=\setparts{\vect{x}\in\complex{n}}{A\vect{x}=\zerovector}$.
First, $\zerovector\in\nsp{A}$, which can be inferred as a consequence of Theorem HSC. So $\nsp{A}\neq\emptyset$.
Second, check additive closure by supposing that $\vect{x}\in\nsp{A}$ and $\vect{y}\in\nsp{A}$. So we know a little something about $\vect{x}$ and $\vect{y}$: $A\vect{x}=\zerovector$ and $A\vect{y}=\zerovector$, and that is all we know. Question: Is $\vect{x}+\vect{y}\in\nsp{A}$? Let us check. \begin{align*} A(\vect{x}+\vect{y})&=A\vect{x}+A\vect{y}&&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=\zerovector+\zerovector&&\vect{x}\in\nsp{A},\ \vect{y}\in\nsp{A}\\ &=\zerovector&&\knowl{./knowls/theorem.VSPCV.knowl}{\text{Theorem VSPCV}} \end{align*} So, yes, $\vect{x}+\vect{y}$ qualifies for membership in $\nsp{A}$.
Third, check scalar multiplication closure by supposing that $\alpha\in\complex{\null}$ and $\vect{x}\in\nsp{A}$. So we know a little something about $\vect{x}$: $A\vect{x}=\zerovector$, and that is all we know. Question: Is $\alpha\vect{x}\in\nsp{A}$? Let us check. \begin{align*} A(\alpha\vect{x})&=\alpha(A\vect{x})&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\alpha\zerovector&&\vect{x}\in\nsp{A}\\ &=\zerovector&&\knowl{./knowls/theorem.ZVSM.knowl}{\text{Theorem ZVSM}} \end{align*} So, yes, $\alpha\vect{x}$ qualifies for membership in $\nsp{A}$.
Having met the three conditions in Theorem TSS we can now say that the null space of a matrix is a subspace (and hence a vector space in its own right!).