Finding a basis for a set of $2\times2$ matrices 
Find a basis for $M_{2\times2}$ containing the matrices $$\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}$$

I know that every $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ can be written as: 
$$a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}  + c \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}  + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$ 
so those matrices are a basis for the vector space of $2\times2$ matrices, but how do I apply this to specific matrices? I know how to find a basis for a set of vectors, but matrices confuse me. 

You can go with more trial & error-like approaches, or go for a more systematic approach. If you can easily add two linearly independent matrices "by inspection", then you are done. This is doable in your case (see Emilio Novati's answer) but can get hard(er) in a more general case.

Knowing the standard basis, you could look at:
$$\left\{
\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix},
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\right\}$$
Because the last 4 matrices form a basis for $M_{2 \times 2}$, this set clearly spans $M_{2 \times 2}$. But since $M_{2 \times 2}$ is 4-dimensional, this set cannot be linearly independent. Work from left to right and omit any matrix which can be written as a linear combination of the matrices before; i.e. deleting the linearly dependent matrices to end up with a basis for $M_{2 \times 2}$ containing the given two matrices.
This can be done in a more systematic way by entering the matrices in columns, identifying the matrix $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ with the column vector $(a,b,c,d)^T$ and then row reducing this matrix; the linearly independent columns (and thus corresponding matrices) will be the pivot columns. This approach is explained in Omnomnomnom's answer as well.

Alternatively, you can check this answer for another systematic approach. I'll leave out the details here, but identifying the matrix $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ with the vector $(a,b,c,d) \in \mathbb{R}^4$, this comes down to finding a basis for the null space of the matrix (WolframAlpha):
$$\begin{pmatrix} 1 & 1 & 2 & 3 \\ 1 & 1 & 3 & 2 \end{pmatrix}$$
So a basis would also be:
$$\left\{
\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix},
\begin{pmatrix} -5 & 0 \\ 1 & 1 \end{pmatrix},
\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}
\right\}$$
