= Solution
Let $T:\mathbb R^2\to D$ be the nearest-point projection,
$$
T(y)=
\begin{cases}
y,&|y|\leq1,\\
y/|y|,&|y|>1.
\end{cases}
$$
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 $|f(x)|\leq1$, so $T(f(x))=f(x)$ and again $f(x)=x$. Thus $f$ has a fixed point.
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