If $f: \mathbb{S} \to \mathbb{S}$ such that $|f(x)-f(y)|≤\frac{|x-y|}{2}$ $\forall ~ x,y \in S$ where $S$ is a finite subset of $\mathbb{R}$, show that $\exists k\in \mathbb{S}$ such that $f(k)=k$.
Why does this seem immensely easy yet not doable? First of all, we can arrange the elements of the set $\mathbb{S}$ in an increasing order, that is, $\mathbb{S}=\{a_i~ |~ a_r<a_s \text{ for } r<s, i \in [1,n]\cap \mathbb{N} \}$. Now, observe that if $|a_i - a_j|$ is minimal for $i=p, j=q$ such that $p\neq q$, then $|f(a_i)-f(a_j)|\leq \frac{|a_i-a_j|}{2}$ is itself a contradiction unless $f(a_i)=f(a_j)$ as otherwise, we will result into obtaining $|a_i-a_j|\leq |f(a_i)-f(a_j)|\leq \frac{1}{2} {|a_i-a_j|}$. (The first inequality is direct from the fact that the function maps from $\mathbb{S}$ to $\mathbb{S}$ and that $|a_i-a_j|$ is minimal.
I couldn't proceed any further.
Just to show an example,
If $\mathbb{S}=\{1,2.1,3.2,4,4.25\}$, then, $f(1)=f(2.1)=f(3.2)=4 ~ ; ~ f(4)=f(4.25)=4.25$ seems to be a valid example!
Suppose $m = \min_x|f(x)-x|= |f(x^*)-x^*|$, then $m \le |f(f(x^*))-f(x^*) | \le {1 \over 2} |f(x^*)-x^*| = {m \over 2}$ and so $m = 0$.