Let $\displaystyle A \in M_{2}(\mathbb(R))$.
If $\displaystyle AB=BA$ for all $\displaystyle B \in M_{2} (\mathbb(R))$,
show that $\displaystyle A=\alpha I_{2}$ for any $\displaystyle \alpha \in \mathbb(R)$.
One way is to write $\displaystyle A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ then select $\displaystyle B=\left[\begin{matrix}1&1\\1&1\end{matrix}\right]$ which gives you a=d and b=c, then select $\displaystyle B=\left[\begin{matrix}1&0\\0&0\end{matrix}\right]$ which gives you c=b=0. (Find AB and BA and set corresponding cells equal.)
If it's true for all B, it's certainly true for the particular B's I mentioned in my first post. Your approach might work but it seems a bit of a mess. Note that you're only asked to prove implication in one direction. Also note that the problem statement is a little imprecise in language, I would write "for some alpha" rather than "for any alpha". (Clearly it's an existential rather than universal statement, because the universal statement makes no sense.)
You might find this useful.
how to make commutative 2X2 matrices