(1) We will use the three criteria of Theorem TSS. The zero vector of $M_{22}$ is the zero matrix, $\zeromatrix$ (Definition ZM), which is a symmetric matrix. So $S_{22}$ is not empty, since $\zeromatrix\in S_{22}$.
Suppose that $A$ and $B$ are two matrices in $S_{22}$. Then we know that $\transpose{A}=A$ and $\transpose{B}=B$. We want to know if $A+B\in S_{22}$, so test $A+B$ for membership,
\begin{align*}
\transpose{\left(A+B\right)}&=\transpose{A}+\transpose{B}&&\knowl{./knowls/theorem.TMA.knowl}{\text{Theorem TMA}}\\
&=A+B&&A,\,B\in S_{22}
\end{align*}
So $A+B$ is symmetric and qualifies for membership in $S_{22}$.
Suppose that $A\in S_{22}$ and $\alpha\in\complex{\null}$. Is $\alpha A\in S_{22}$? We know that $\transpose{A}=A$. Now check that,
\begin{align*}
\transpose{\alpha A}&=\alpha\transpose{A}&&\knowl{./knowls/theorem.TMSM.knowl}{\text{Theorem TMSM}}\\
&=\alpha A&&A\in S_{22}
\end{align*}
So $\alpha A$ is also symmetric and qualifies for membership in $S_{22}$.
With the three criteria of Theorem TSS fulfilled, we see that $S_{22}$ is a subspace of $M_{22}$.
(2) An arbitrary matrix from $S_{22}$ can be written as
$\begin{bmatrix}
a&b\\b&d
\end{bmatrix}$.
We can express this matrix as
\begin{align*}
\begin{bmatrix}
a&b\\b&d
\end{bmatrix}
&=
\begin{bmatrix}
a&0\\0&0
\end{bmatrix}+
\begin{bmatrix}
0&b\\b&0
\end{bmatrix}+
\begin{bmatrix}
0&0\\0&d
\end{bmatrix}\\
&=
a
\begin{bmatrix}
1&0\\0&0
\end{bmatrix}+
b
\begin{bmatrix}
0&1\\1&0
\end{bmatrix}+
d
\begin{bmatrix}
0&0\\0&1
\end{bmatrix}
\end{align*}
this equation says that the set
\begin{equation*}
T=\set{
\begin{bmatrix}
1&0\\0&0
\end{bmatrix},\,
\begin{bmatrix}
0&1\\1&0
\end{bmatrix},\,
\begin{bmatrix}
0&0\\0&1
\end{bmatrix}
}
\end{equation*}
spans $S_{22}$. Is it also linearly independent?
Write a relation of linear dependence on $S$,
\begin{align*}
\zeromatrix&=
a_1
\begin{bmatrix}
1&0\\0&0
\end{bmatrix}+
a_2
\begin{bmatrix}
0&1\\1&0
\end{bmatrix}+
a_3
\begin{bmatrix}
0&0\\0&1
\end{bmatrix}\\
\begin{bmatrix}
0&0\\0&0
\end{bmatrix}
&=
\begin{bmatrix}
a_1&a_2\\a_2&a_3
\end{bmatrix}
\end{align*}
The equality of these two matrices (Definition ME) tells us that $a_1=a_2=a_3=0$, and the only relation of linear dependence on $T$ is trivial. So $T$ is linearly independent, and hence is a basis of $S_{22}$.
(3) The basis $T$ found in part (2) has size 3. So by Definition D, $\dimension{S_{22}}=3$.