Let $(K,d)$ be a compact metric space. We consider $f: K -> K$ : $$\forall x,y \in K: d(x,y) \leq d(f(x), f(y))$$
Show that $$\forall \epsilon >0 :d(f(x), f(y)) \leq d(x, y)+\epsilon$$
What I've tried to do was following: $$d(f(x), f(y)) \leq d(f(x),x) + d(x,y) + d(f(y),y)$$ Now I need to show that $$\forall \epsilon >0 :d(f(x), x) \leq \epsilon$$ Let $D(x):=d(x,f(x))$ - compact-continuous mapping => uniformly continuous (Heine–Cantor theorem) - how can I use that though? Let $D(x)$ reach its maximum at $x_0$, so$$D(f(x_0))=d(f(f(x_0)),f(x_0)) \geq d(f(x_0),x_0)=D(x_0)=max(D(x))=>$$ $$=>D(x_0)=D(f(x_0))=...=D(f^n(x_0))=...$$ Can I somehow condlude that $$\forall \epsilon >0: D(x_0) \leq \epsilon$$?
What bothers me is that I haven't used much the fact that K is a compact - although I don't see any ways to use it. Perhaps I should consider 2 sequences $$\{x, f(x), f(f(x)),...\}, \{y, f(y), f(f(y)),...\}$$ and show that their subsequential limits are equal to $x, y$ respectively.
Thanks in advance.
Your approach cannot work, because that would imply $d(f(x),x)=0$, and hence $f=\text{id}_K$, but clearly if $K=[-1,1]$ say, then $f(x)=-x$ is also an isometry. So, we need another approach. Also, in your assumptions it is not stated that $f$ is continuous, yet you have invoked continuity. I think whenever it comes to questions about contractions/expansions, it is always a good idea to consider iterations of a given point as in your last paragraph.
Here's a proof which I think is low-tech and straight forward, but a little messy because we have to repeatedly take subsequences to make everything work out. First of all we make an observation: given $x\in K$, let's define $x_n:=f^{(n)}(x)$. By compactness, there is a convergent subsequence $\{x_{n_j}\}_{j=1}^{\infty}$. By passing to another subsequence, we may as well assume $n_2-n_1<n_3-n_2<\dots$. In other words, if we define $m_j=n_{j+1}-n_j$, then $\{x_{m_j}\}_{j=1}^{\infty}$ is a subseqence of $\{x_n\}_{n=1}^{\infty}$. Now, the nice thing is for every $j\in \Bbb{N}$, \begin{align} d(x_{m_j},x)\leq d(f^{n_j}(x_{m_j}),f^{n_j}(x))=d(x_{n_{j+1}},x_{n_j}) \end{align} and the RHS approaches $0$ as $j\to \infty$ ($x_{n_j}$ and $x_{n_{j+1}}$ have the same limit; another way of saying it is that $\{x_{n_j}\}_{j=1}^{\infty}$ is a Cauchy sequence).
Now, let $x,y\in K$ be arbitrary. By what we have shown above, $\{x_n=f^{(n)}(x)\}_{n=1}^{\infty}$ has a subsequence $\{x_{m_j}\}_{j=1}^{\infty}$ such that $x_{m_j}\to x$. Now, by similar reasoning, the sequence $\{y_{m_j}=f^{(m_j)}(y)\}_{j=1}^{\infty}$ also has a subsequence $\{y_{m_{j_l}}\}_{l=1}^{\infty}$ converging to $y$. Therefore, for any $l\in\Bbb{N}$, we have by the expansion property, \begin{align} d(f(x),f(y))\leq d(x_{m_{j_l}},y_{m_{j_l}}), \end{align} and the RHS approaches $d(x,y)$. Thus, we have shown $d(x,y)=d(f(x),f(y))$.
In fact, we can actually go one step further and prove $f$ is bijective (injectivity is obvious from the expansion), for surjectivity, note that the first part of our proof shows $f(K)$ is dense in $K$, but now $f$ being an isometry implies $f$ is continuous, hence $f(K)$ is compact hence closed, hence $K=\overline{f(K)}=f(K)$, which shows surjectivity.