I want to show that if $f$ is continuous and $\vert f(x) - f(f(x))\vert \leq \frac{1}{2}\vert f(x) - x \vert$, that there exists an $x_0$ such that $f(x_0) = x_0$ without using any sequence $x_n$ (since the point of this question is to minimize the distance between $f(x_0)$ and $x_0$).
Say $f(x_0) = y$, and so we have $\vert y - f(y)\vert \leq \frac{1}{2}\vert y-x_0 \vert$. Now can we use continuity to somehow make this less than $\epsilon$ somehow?
Another ideaI had was to consider like some $\vert f^n(x) - f^{n+1}(x)\vert \leq \frac{1}{2^n}\vert f(x) - x\vert$ but I don't know how to use the continuity of f. Any direction would be nice.
I don't know what you mean by ' without using any sequence $\{x_n\}$ but here is a proof: $|f^{n}-f^{n+m}| \leq |f^{n}-f^{n+1}|+|f^{n+1}-f^{n+2}|+...+|f^{n+m-1}-f^{n+m}| \leq (\sum_{k=n}^{n+m-1} \frac 1 {2^k}|f(x)-x| $. From the convergence of the series $\sum_1^{\infty} \frac 1 {2^k}$ it follows that the sequence $\{f^{n} (x)\}$ is Cauchy, hence convergent. If $a=\lim_{n\to \infty} f^{n}(x)$ the continuity of $f$ gives $f(a)=a$.