We want to show that $W=\spn{S}$ (Definition SSVS), which is an equality of sets (Definition SE).
First, show that $\spn{S}\subseteq W$. Begin by checking that each of the three matrices in $S$ is a member of the set $W$. Then, since $W$ is a vector space, the closure properties (Property AC, Property SC) guarantee that every linear combination of elements of $S$ remains in $W$.
Second, show that $W\subseteq\spn{S}$. We want to convince ourselves that an arbitrary element of $W$ is a linear combination of elements of $S$. Choose
\begin{equation*}
\vect{x}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\in W
\end{equation*}
The values of $a,\,b,\,c,\,d$ are not totally arbitrary, since membership in $W$ requires that
$2a-3b+4c-d=0$. Now, rewrite as follows,
\begin{align*}
\vect{x}
&=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\
&=\begin{bmatrix}a&b\\c&2a-3b+4c\end{bmatrix}
&&2a-3b+4c-d=0\\
&=
\begin{bmatrix}a&0\\0&2a\end{bmatrix}+
\begin{bmatrix}0&b\\0&-3b\end{bmatrix}+
\begin{bmatrix}0&0\\c&4c\end{bmatrix}
&&\knowl{./knowls/definition.MA.knowl}{\text{Definition MA}}\\
&=
a\begin{bmatrix}1&0\\0&2\end{bmatrix}+
b\begin{bmatrix}0&1\\0&-3\end{bmatrix}+
c\begin{bmatrix}0&0\\1&4\end{bmatrix}
&&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\
&\in\spn{S}
&&\knowl{./knowls/definition.SS.knowl}{\text{Definition SS}}
\end{align*}