Let $(M,d)$ be a compact metric space and $f:M \to M$ such that $d(f(x),f(y)) \ge d(x,y) , \forall x,y \in M$ , then $f$ is isometry?

Let $a,b$ be two elements of $M$. We wish to show that $d(f(a),f(b)) \leq d(a,b)$.

As noted in the OP, there is an increasing sequence $(r_n)$ of integers such that $f^{r_n}(a)$ converges in $M$ and $f^{r_n}(b)$ converges in $M$. Then $d(f^{r_n}(a),f^{r_{n+1}}(a))$ tends to zero. But $d(a,f^{r_{n+1}-r_n}(a)) \leq d(f^{r_n}(a),f^{r_{n+1}}(a))$ by the hypothesis on $f$, so $d(a,f^{r_{n+1}-r_n}(a))$ tends to zero. For the same reason, $d(b,f^{r_{n+1}-r_n}(b))$ tends to zero. There are then two not-so-different cases to consider :

First case : the set of values $V=\lbrace r_{n+1}-r_n\rbrace$ is finite. Then some value $v\in V$ is attained infinitely often ; it follows that $d(a,f^v(a))=0$ hence $f^v(a)=a$ and similarly $f^v(b)=b$. If we put $x=f(a)$ and $y=f(b)$, we have $d(x,y) \leq d(f^{v-1}(x),f^{v-1}(y))$ by the hypothesis on $f$. But this is exactly what we want.

Second case : the set of values $V=\lbrace r_{n+1}-r_n\rbrace$ is infinite. Then there is an increasing subsequence $(s_n)$ of $(r_{n+1}-r_n)$ such that $f^{s_n}(a)$ converges to $a$ and $f^{s_n}(b)$ converges to $b$. Then $u_n=d(f^{s_n}(a),f^{s_n}(b))$ converges to $d(a,b)$. On the other hand, by the hypothesis on $f$ we have

$$ d(a,b)\leq d(f(a),f(b)) \leq u_0 \leq u_1 \leq \ldots \leq d(a,b) $$

this concludes the proof.

I would proceed like this :

1) $\overline{f(M)} = M$. Indeed, as $M$ is compact, if $a\in M$, the sequence $(f^n (a))_{n\in\mathbf{N}}$ has a convergent subsequence, which is therefore Cauchy : this means that for all $\varepsilon$ you can find $n,p\in\mathbf{N}$ such that $d(f^n (a),f^{n+p}(a))< \varepsilon$, and the hypothesis on $f$ implies then that $d(a,f^p(a))<\varepsilon$, and this shows that $a\in \overline{f(M)}$.

2) As $(M,d)$ is compact, so is $(M\times M, d\times d)$ where $$(d\times d)((m_1,m_2),(m'_1,m'_2)) := d(m_1,m'_1)+d(m_2,m'_2)$$ for all $m_1,m'_1,m_2,m'_2\in M$. Moreover, the application $g=(f,f) : M\times M \to M\times M$ verifies the same property as the one verified by $f$. This implies that, if $x,y\in M$ we have : $$\forall \varepsilon > 0, \exists p\in\mathbf{N}, d(x,f^p(x))<\varepsilon\textrm{ and }d(y,f^p(y))<\varepsilon.$$

3) Now suppose having $x,y\in M$ such that $d(f(x),f(y)) > d(x,y)$. Then choose $\varepsilon > 0$ such that $$(A)\;\;\;\;\;\;\;\;d(x,y) + 2\varepsilon < d(f(x),f(y)).$$ Then for all $k\in\mathbf{N}^*$ we have $$(F)\;\;\;\;\;\;\;\;d(f^p(x),f^p(y)) \geq d(f(x),f(y)) > d(x,y).$$ But then 2 implies that we can find a $p$ such that $$(B)\;\;\;\;\;\;\;\;d(x,f^p(x))<\varepsilon\textrm{ and }d(y,f^p(y))<\varepsilon.$$ Then (A), (B) and the triangle inequality imply that $$d(f^p(x),f^p(y)) < d(x,y) + 2\varepsilon < d(f^p(x),f^p(y))$$ which is in contradiction with (F).

Remark. Note that now that $f$ is an isometry, $f$ is continuous so that $f(M)$ is compact (i.e. quasi-compact and Hausdorff) and is therefore closed, so that 1 implies that $f$ is in fact surjective.