Here is a different kind of answer that is just a lot harder for no reason, but might help you transition.
Suppose we had a specific B, such as: $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$ We want to find all A so that AB = BA. Well, we don't know the entries of A, so we replace the unknown numbers with... well unkowns, also known as variables. $A = \begin{bmatrix} x & y \\ z & t \end{bmatrix}$
Ok, now we write down what we know about the variables from AB = BA: $ AB = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 2x+4y & 3x+5y \\ 2z+4t & 3z+5t \end{bmatrix} $ $ BA = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} 2x+3z & 2y+3t \\ 4x+5z & 4y+5t \end{bmatrix} $ For two matrices to be equal, we need to have the entries equal: $\left\{\begin{align} 2x + 4y &= 2x + 3z \\ 2z+4t &= 4x+5z \\ 4x+5y &= 2y + 3t \\ 3z+5t &= 4y+5t \end{align}\right.$ What do you know? We have a system of linear equations. We can solve them as usual: find a particular solution (oooo let me let me!): $(x=0,y=0,z=0,t=0)$. Now we know all other solutions are found by adding a "homogeneous solution", which form a subspace.
If we want, we can be even more linear-algebra-y. let's move all the variables to one side:
$\left\{\begin{align} 0x + 4y - 3z + 0t &= 0 \\ -4x+0y-3z+4t &= 0 \\ 4x+3y +0z -3t &= 0 \\ 0x-4y+3z+0t &= 0 \end{align}\right.$
We can write it as a matrix equation: $\left[\begin{array}{rrrr} 0 & 4 & -3 & 0 \\ -4 & 0 & -3 & 4 \\ 4 & 3 & 0 & -3 \\ 0 & -4 & 3 & 0 \end{array}\right] \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
So we are just asking for the null space of the matrix! Clearly that is a subspace. We could even do Gaussian elimination to find which one. I won't.