= Solution
Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The <contraction mapping> estimate
$$
d(x_{n+1},x_n)\leq\lambda^nd(x_1,x_0)
$$
makes $(x_n)$ Cauchy by a <geometric series>. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\leq\lambda d(x_0,y_0)$, so they coincide. This proves the <Banach fixed-point theorem>.
Now suppose the final assertion failed. There would be $m_k,n_k\to\infty$ with $f(a_{m_k})=a_{n_k}$. Since $a_n\to a$ and $f$ is continuous, the two sides tend respectively to $f(a)$ and $a$, so $f(a)=a$. Uniqueness would give $a=x_0$, a contradiction. Hence the required \b[$\ell$ and $m$ exist].
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