please how to solve this question
Let $E$ a compact metric space and a function $f:E\to E$ such that $$ \forall x,y \in E, d(f(x),f(y))\geq d(x,y)$$ Let $a\in E$ and a sequence $(f^n(a))$ , prove that $a$ in an adherent value of $(f^n(a))$.
As E is compact then $(f^n(a))$ has a convergent subsequence, but I stop here
Hint: Prove by contradiction. Suppose $(f^n(a))$ stays away from $a$, so there is some $\varepsilon>0$ such that $d(f^n(a),a)>\varepsilon$ for all $n\geq 1$. Prove that $(f^n(a))_{n>m}$ also stays away from $f^m(a)$ for all $m$ (with the same $\varepsilon$), and show this yields a contradiction with compactness of $E$.