 Solution

Let $T:{\mathbb{R}}^2\to D$ be the nearest-point projection,It is continuous. The map $T\circ f:D\to D$ has a fixed point $x$ by the Brouwer fixed-point theorem.

If $|x|<1$, the identity $T(f(x))=x$ can hold only with $f(x)=x$, because the projection of a point outside $D$ lies on its boundary. If $|x|=1$, the hypothesis gives $|f(x)|\le 1$, so $T(f(x))=f(x)$ and again $f(x)=x$. Thus $f$ has a fixed point.