$\text{Let $(X,d) $ be a compact metric space and $f:X\longrightarrow X$ a continuous operator}$
$\text{such that } d\big(f(x),f(y)\big)<d(x,y)\text{ } \forall x,y\in X, x\neq y.$
$\text{I know that there exists a unique fixed point $x_0\in X: f(x_0)=x_0$},$
$\text{however, how can we show that: }$
$$\lim_{n\to\infty}f^n(x)=x_0 \text{ }\forall x\in X$$
$\text{Where }f^n(x):=f\color{black}{\underbrace{ ofofo\dots o f(x) }_{ \text{$n$ times } }}.$
$$\bullet \text{ My approach: $X$ is compact, hence $\exists$ subsequence $f^{k_n}(x) :k_n\ge n$ s.t}$$ $$f^{k_n}(x) \longrightarrow a\in X,\text{ $f$ is continuous, thus $f^{k_n +1}(x)\longrightarrow f(a)\implies f(a)=a\implies a=x_0$}$$
$\text{How can i continue from this point on, any ideas?}$
You're almost there: To finish you can use the Uryshon subsequence principle which can be found here: https://www.math.ucla.edu/~hmkhang24/131BHW19/Notes.pdf
Indeed for any $x \in X$ take the sequence $(f^n(x))_k$. Take a subsequence $(f^n_k(x))_k$. Then with this subsequence you use your argument (compactness $\implies$ convergent subsequence which converges to $x_0$). Then use Uryshon's subsequence principle (also called subsubsequence principle) since you took an arbitrary subsequence of the original sequence.
Alternatively you can argue by contradiction (which amounts to using the subsequence principle). Suppose your sequence does not converge to $x_0$, then there is a subsequence which does not converge to $x_0$, for some $x \in X$. But this subsequence has a convergent subsequence due to compactness and then you use the argument that you gave.