My Try:
$|f'(a)|>1$.
Assume that the sequence converges to a limit $b$. Then $f(b)=b$. Since $a$ is the only fixed point it implies that $b=a$.
Hence, given any $\frac{1}{m}$ where $m\in \mathbb{N}$ there is $N$ such that $|x_n-a|<\frac{1}{m}$ for all $n>N$. Then $|x_{N+1}-a|<\frac{1}{m}$ and $|x_{N+2}-a|<\frac{1}{m}$. So, $|f(x_{N+1})-f(a)|<\frac{1}{m}$.
Since $f$ is strictly decreasing, $f(a-\frac{1}{m})>f(x_{N+1})> f(a+\frac{1}{m})$. So we have $f(a+\frac{1}{m})-f(a)<\frac{1}{m}$ and $f(a-\frac{1}{m})-f(a)>-\frac{1}{m}$.
I wanted to get a contradiction but now stuck in here. Can anybody please help me to complete? Or am I on the wrong track?
If this sequence $(x_n)$ is convergent, then its limit is $a$ (as you argued). Note that, since $x_n\to a$, we must have $$ f'(a) = \lim_{n\to \infty}\frac{f(x_n)-a}{x_n - a}. $$
Since $f'(a)<-1$, there must be a $N>0$ such that, if $n>N$, $$ \frac{f(x_n)-f(a)}{x_n-a}<-1 \implies \left|\frac{f(x_n)-f(a)}{x_n-a} \right| > 1 \implies \left|x_{n+1}-a \right|>\left|x_n -a \right|. $$
Then, in particular, $|x_N-a|<|x_{N+1}-a|<\ldots$, which is a contradiction, because the sequence is convergent.