The question is simple: Suppose $f : \mathbb{R}^ 2 \to \mathbb{R}^ 2$ is continuous. Show that there exist $\lambda > 0$ and $x \in \mathbb{R}^2$ such that $f(x) = \lambda x$.
So basically, we have to show that there is at least one point that gets scaled by $f$. If $f$ has a fixed point $x^*$, then we are done, since we can then take $x = x^ *$ and $\lambda = 1$ and we are done. What we could do is arrive at a contradiction when assuming that the following is not true: for all $\lambda > 0$, we have that $(1/\lambda) f(x)$ has no fixed point $x^*$. Because then, by contradiction, there is a $\lambda$ such that the function $(1/\lambda) f(x)$ has a fixed point $x^*$, so that $f$ had to scale $x^*$ by exactly $\lambda$, which is what we wanted.
However, why should we arrive at contradiction? Since $f$ has no fixed point, why is it a problem if $(1/\lambda) f$ doesn't have one either?