By Definition SS, the span contains linear combinations of vectors from the vector space $V$, so by repeated use of the closure properties, Property AC and Property SC, $\spn{S}$ can be seen to be a subset of $V$.
We will then verify the three conditions of Theorem TSS. First, \begin{align*} \zerovector &=\zerovector+\zerovector+\zerovector+\ldots+\zerovector&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\ &=0\vect{u}_1+0\vect{u}_2+0\vect{u}_3+\cdots+0\vect{u}_t&&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}} \end{align*}
So we have written $\zerovector$ as a linear combination of the vectors in $S$ and by Definition SS$, \zerovector\in\spn{S}$ and therefore $\spn{S}\neq\emptyset$.
Second, suppose $\vect{x}\in\spn{S}$ and $\vect{y}\in\spn{S}$. Can we conclude that $\vect{x}+\vect{y}\in\spn{S}$? What do we know about $\vect{x}$ and $\vect{y}$ by virtue of their membership in $\spn{S}$? There must be scalars from $\complex{\null}$, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_t$ and $\beta_1,\,\beta_2,\,\beta_3,\,\ldots,\,\beta_t$ so that \begin{align*} \vect{x}&=\lincombo{\alpha}{u}{t}\\ \vect{y}&=\lincombo{\beta}{u}{t} \end{align*} Then \begin{align*} \vect{x}+\vect{y}&=\lincombo{\alpha}{u}{t}\\ &\quad\quad+\lincombo{\beta}{u}{t}\\ &=\alpha_1\vect{u}_1+\beta_1\vect{u}_1+\alpha_2\vect{u}_2+\beta_2\vect{u}_2\\ &\quad\quad+\alpha_3\vect{u}_3+\beta_3\vect{u}_3+\cdots+\alpha_t\vect{u}_t+\beta_t\vect{u}_t&&\knowl{./knowls/property.AA.knowl}{\text{Property AA}}, \knowl{./knowls/property.C.knowl}{\text{Property C}}\\ &=(\alpha_1+\beta_1)\vect{u}_1+(\alpha_2+\beta_2)\vect{u}_2\\ &\quad\quad+(\alpha_3+\beta_3)\vect{u}_3+\cdots+(\alpha_t+\beta_t)\vect{u}_t&&\knowl{./knowls/property.DSA.knowl}{\text{Property DSA}} \end{align*} Since each $\alpha_i+\beta_i$ is again a scalar from $\complex{\null}$ we have expressed the vector sum $\vect{x}+\vect{y}$ as a linear combination of the vectors from $S$, and therefore by Definition SS we can say that $\vect{x}+\vect{y}\in\spn{S}$.
Third, suppose $\alpha\in\complex{\null}$ and $\vect{x}\in\spn{S}$. Can we conclude that $\alpha\vect{x}\in\spn{S}$? What do we know about $\vect{x}$ by virtue of its membership in $\spn{S}$? There must be scalars from $\complex{\null}$, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_t$ so that \begin{align*} \vect{x}&=\lincombo{\alpha}{u}{t}\\ \end{align*} Then \begin{align*} \alpha\vect{x}&=\alpha\left(\lincombo{\alpha}{u}{t}\right)\\ &=\alpha(\alpha_1\vect{u}_1)+\alpha(\alpha_2\vect{u}_2)+\alpha(\alpha_3\vect{u}_3)+\cdots+\alpha(\alpha_t\vect{u}_t)&&\knowl{./knowls/property.DVA.knowl}{\text{Property DVA}}\\ &=(\alpha\alpha_1)\vect{u}_1+(\alpha\alpha_2)\vect{u}_2+(\alpha\alpha_3)\vect{u}_3+\cdots+(\alpha\alpha_t)\vect{u}_t&&\knowl{./knowls/property.SMA.knowl}{\text{Property SMA}}\\ \end{align*} Since each $\alpha\alpha_i$ is again a scalar from $\complex{\null}$ we have expressed the scalar multiple $\alpha\vect{x}$ as a linear combination of the vectors from $S$, and therefore by Definition SS we can say that $\alpha\vect{x}\in\spn{S}$.
With the three conditions of Theorem TSS met, we can say that $\spn{S}$ is a subspace (and so is also a vector space, Definition VS). (See Exercise SS.T20, Exercise SS.T21, Exercise SS.T22.)