Answer
See below
Work Step by Step
$A=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}$ is a skew symmetric.
Let $S=\{A \in M_2(R)\}$
Since $A \in S$ we have $A^T=-A\rightarrow \begin{bmatrix}
a & b \\ c & d
\end{bmatrix}=\begin{bmatrix}
-a & -b \\ -c & -d
\end{bmatrix}$
Thus, $a=-a\\
b=-b\\
c=-c\\
d=-d$
then $a=d=0\\
c=-b$
Obtain: $\begin{bmatrix}
0 & b \\ -b & 0
\end{bmatrix}=b\begin{bmatrix}
0 & 1 \\ -1 & 0
\end{bmatrix}$
We can see that every matrix $A \in S$ is a scalar multiple of the matrix $\begin{bmatrix}
0 & 1\\ -1 & 0
\end{bmatrix}$
Hence, the set $\{\begin{bmatrix}
0 & 1\\ -1 & 0
\end{bmatrix}\}$ spans $S$
