Fix $x\in X$ and let $x_n=f^n(x)$  ,$n=1,2,...$ yet again we break the argument into two steps.

step 1:$\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=0$

Since $f$ is contractive the sequence $\{d(x_n,x_{n+1})\}$ is monotone decreasing and bounded below  ,so $$\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=r\geq0$$. Assume that $r>0$.Then by the contractive condition : $$\frac{d(x_{n+1},x_{n+2})}{d(x_n,x_{n+1})}\leq \alpha(d(x_n,x_{n+1}))\,\,\,,n=1,2,3...$$ $n\rightarrow\infty\implies 1\leq\lim\limits _{n\rightarrow\infty}\alpha(d(x_n,x_{n+1}))\implies r=0$ contradiction.

Fix $x\in X$ and let $x_n=f^n(x)$, $n=1,2, \ldots$ yet again we break the argument into two steps.

Step 1:  $\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=0$

Since $f$ is contractive the sequence $\{d(x_n,x_{n+1})\}$ is monotone decreasing and bounded below, so $$\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=r\geq0$$ Assume that $r>0$. Then by the contractive condition:

$$\frac{d(x_{n+1},x_{n+2})}{d(x_n,x_{n+1})}\leq \alpha(d(x_n,x_{n+1}))\,\,\,,n=1,2,3, \ldots$$

Then $n\rightarrow\infty\implies 1\leq\lim\limits _{n\rightarrow\infty}\alpha(d(x_n,x_{n+1}))\implies r=0$, contradiction.

Step 2: $\{x_n\}$ is a Cauchy sequence:

Assume $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$.

$$d(x_n,x_m)\leq d(x_n,x_{n+1})+d(x_{n+1},x_{m+1})+d(x_{m+1},x_m)$$ So by the contractive condition : $$d(x_n,x_m)\leq {(1-\alpha d(x_n,x_{m}))}^{-1}[d(x_{n},x_{n+1})+d(x_{m+1},x_m)]$$ Under the assumption $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$,step 1 implies $\lim\limits_{n,m\rightarrow\infty}\sup {(1-\alpha (d(x_n,x_m)))}^{-1}=+\infty$

From which $\lim\limits_{n,m\rightarrow\infty}\sup \alpha (d(x_n,x_m))=1$, that implies $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)=0$, again a contradiction.