Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$.

The eigenvalues of $A$ (or $D$) satisfy $|\lambda | = { 1\over \sqrt{2}} <1$. Hence $|\lambda|^n \to 0$.

While the answer provided by @cooper.hat is probably the one you want (as it finishes your otherwise correct proof), I would like to offer a different approach.

If $A^2=A-(1/2)I$, then multiplying by $A$ repeatedly and using induction, we see that $A^n=a_n A +b_nI$, and further that $A^{n+1}=a_n A^2 +b_n A=(a_n+b_n)A-(1/2)a_nI.$ Looking at the first few $(a_n,b_n)$ pairs, we have $$(1,0), (1,-1/2), (1/2,-1/2),(0,-1/4)$$

So $A^{4}=(-1/4)I$. This makes it easy to compute larger powers of $A$ and see they are tending towards $0$.

Let's do a few computations. We have:

$$A^2=A-{1\over 2}I$$

This means $A^3={A\over 2}-{I\over 2}$ and:

$$A^4=A^2-A+{1\over 4}I=-{1\over 4}I$$

We can prove by induction that

$$\begin{align} A^{4k}&=\left({-1\over 4k}\right)^k\cdot I\\ A^{4k+1}&=\left({-1\over 4k}\right)^k\cdot A\\ A^{4k+2}&=\left({-1\over 4k}\right)^k\cdot\left(A-{1\over 2}I\right)\\ A^{4k+3}&=\left({-1\over 4k}\right)^k\cdot\left({A\over 2}-{1\over 2}I\right) \end{align}$$

Let

$$M=\sup\{1,\|A\|,\|A-{I\over 2}\|,\|{A\over 2}-{I\over 2}\|\}$$

We have

$$\|A^n\|\leq {M\over n^{\lfloor {n\over 4}\rfloor}}\to 0$$