**(a) **It's enough to prove the following:

If there exists $\displaystyle r,s \in K$ such that $\displaystyle d(f(r),f(s))>d(r,s)$, then there exists $\displaystyle t,u \in K$ such that $\displaystyle d(f(t),f(u))<d(t,u)$.

**Proof: **Suppose not and $\displaystyle \forall x,y \in K

$, $\displaystyle d(f(x),f(y)) \ge d(x,y).$ Define a sequence $\displaystyle (f_n)$ as follows:

$\displaystyle f_1=f$

$\displaystyle f_2=f\circ f$

$\displaystyle \vdots$

$\displaystyle f_n=f \circ f_{n-1}$

$\displaystyle \vdots$

Since we have for $\displaystyle x,y \in K$, $\displaystyle d(f(x),f(y)) \ge d(x,y)$, then $\displaystyle d(f_n(x),f_n(y)) \ge d(x,y).$ Since $\displaystyle K$ is a compact metric space, it is sequentially compact, so the sequence $\displaystyle (f_n(x))$ has a Cauchy subsequence $\displaystyle (f_{n_j}(x))$. So for $\displaystyle \varepsilon > 0$ pick $\displaystyle m$ such that for $\displaystyle k \ge 1$

$\displaystyle d(f_{n_{m+k}}(x),f_{n_m}(x)) < \frac{\varepsilon}{2}$

$\displaystyle d(f_{n_{m+k}}(y),f_{n_m}(y)) < \frac{\varepsilon}{2}$.

Then we 'go back' $\displaystyle n_m$ times to get

$\displaystyle d(f_{n_{m+k}-n_m}(x),x) \le d(f_{n_{m+k}}(x),f_{n_m}(x)) <\frac{\varepsilon}{2}$

$\displaystyle d(f_{n_{m+k}-n_m}(y),y) \le d(f_{n_{m+k}}(y),f_{n_m}(y)) <\frac{\varepsilon}{2}$.

Therefore,

$\displaystyle d(f(x),f(y)) \le d(f_{n_{m+k}-n_m}(x),f_{n_{m+k}-n_m}(y))$

$\displaystyle \le d(x,y)+d(f_{n_{m+k}-n_m}(x),x)+d(f_{n_{m+k}-n_m}(y),y)$

$\displaystyle <d(x,y)+\varepsilon $,

and since $\displaystyle \varepsilon$ is arbitrary, $\displaystyle d(f(x),f(y))< d(x,y)$, a contradiction.

**(b)** Since $\displaystyle f$ is an isometry, it is clearly injective and continuous. If $\displaystyle C \subseteq K$ is closed, then it is compact. Therefore, $\displaystyle f(C)$ is compact, which implies that $\displaystyle f(C)$ is closed ($\displaystyle K$ is Hausdorff).

Suppose $\displaystyle f$ is not surjective. Then let $\displaystyle x_0 \in K-f(K)$. Since $\displaystyle f(K)$ is closed, $\displaystyle K-f(K)$ is open. Then for $\displaystyle \varepsilon >0, \, \, B_{\varepsilon}(x_0) \subseteq K-f(K)$, so for $\displaystyle x \in f(K)$, $\displaystyle d(x_0,x) \ge \varepsilon$.

Define a sequence as follows:

$\displaystyle x_1=f(x_0)$

$\displaystyle x_2=f(x_1)$

$\displaystyle \vdots$

$\displaystyle x_n=f(x_{n-1})$

$\displaystyle \vdots$

and let $\displaystyle X_n=\{x_1,x_2,\dots , x_n\}.$ I claim that $\displaystyle d(x_i,x_j) \ge \varepsilon$ for $\displaystyle i \not= j$. Prove by induction.

Since $\displaystyle x_1 \in f(K)$, $\displaystyle d(x_0,x_1) \ge \varepsilon$. Suppose it's true for the elements of $\displaystyle X_n$ for $\displaystyle n \ge 1$. We must show that $\displaystyle d(x_i,x_{n+1}) \ge \varepsilon$ where $\displaystyle x_i \in X_n$. It's true for $\displaystyle i=0$ since $\displaystyle x_{n+1} \in f(K)$. If $\displaystyle i>0$, then $\displaystyle d(x_i,x_{n+1})=d(f(x_{i-1}),f(x_n))=d(x_{i-1},x_n) \ge \varepsilon$ since $\displaystyle x_{i-1},x_n \in X_n$. Let

$\displaystyle X=\bigcup_{n \in \mathbb{N}}X_n$.

Then $\displaystyle X$ is a discrete set. Since $\displaystyle X$ is an infinite set with no limit point, $\displaystyle K$ is not limit point compact, a contradiction.