 Solution

Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The contraction mapping estimatemakes $(x_n)$ Cauchy by a geometric series. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\le \lambda d(x_0,y_0)$, 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 $\ell$ and $m$ exist.