Fixed point in compact metric space 
Let $(X,d)$ be a compact metric space and let $T:X\longrightarrow X$ be a map such that $d(T(x),T(y))<d(x,y)$ for all $x,y\in X$ such that $x\neq y.$
(a) Prove that $T$ has a fixed point  
(b) Prove that the fixed point is obtained as a limit of $\{T^{k}(x_{0})\}$ for 
      $x_{0}\in X$

I have solved (a), however, I'm not able to prove (b).
In particular, I am having some problems using the compactness hypothesis on $X$ to find proper subsequences which are convergent.
 A: The following is probably not the most elegant solution, but nevertheless works.
Have a look at this nice answer https://math.stackexchange.com/a/409832/151552.
Observe that for an arbitrary subsequence $T^{n_k}(x_0)$, we have $T^{n_{k_\ell}}(x_0) \to x$ for some $x \in K$ and a suitable sub(-sub-)sequence (by compactness).
But $T^{n_{k_\ell}}(x_0) \in T^{n_{k_\ell}}(X) \subset \bigcap_{m=1}^N T^m (X) =: K_N$ as soon as $n_{k_\ell} \geq N$. Hence, $x \in \overline{K_N} = K_N$ (note that $K_N$ is closed).
As $N \in \Bbb{N}$ was arbitrary, $x \in \bigcap_N K_N = A$, where $A$ is chosen as in the answer linked above.
But the answer linked above shows that $A = \{\xi_0\}$, where $\xi_0$ is the unique fixed point of $T$. Hence, $x=\xi_0$.
Now use the fact that $(T^{n_k} x_0)_k$ was an arbitrary subsequence so conclude $T^n x_0 \to \xi_0$ (otherwise, there is $\varepsilon > 0$ and a subsequence $(T^{n_k}x_0)_k$ with $d(T^{n_k} x_0, \xi_0) \geq \varepsilon$ for all $k$, which contradicts the convergence $T^{n_{k_\ell}} x_0 \to \xi_0$ shown above).
A: I would love to assume that there is no limit, then there would be at least two adherent values (due to compactness and continuity and infinity), then I may find contradiction somewhere.
