Banach fixed point problem and dependence of the fixed point by a parameter. Let $(X,d)$ be a complete metric space and let $(\Lambda,d_\Lambda)$ be a metric space. Suppose we have a continuous map $F:X \times \Lambda \to X$ such that there exist $0<k<1$ satisfying $$d(F(x,\lambda),F(y,\lambda)) \leq k d(x,y)$$
for all $x,y \in X$ and all $\lambda \in \Lambda$.
By the Banach fixed point for each $\lambda$ there is a unique $x(\lambda)\in X$ satisfying $F(x(\lambda),\lambda) = x(\lambda)$. 
I have to prove that the map $\Lambda \ni \lambda \to x(\lambda)\in X$ is continuous. How do I prove this map is continuous?
$\textbf{My attempt:}$
If we define $f_\lambda(x) = F(x,\lambda)$ we have that for an arbitrary point $p\in X$ the identity $$x(\lambda) = \lim_{n\to \infty} f_\lambda^n(p)$$ holds by the Banach fixed point. Therefore:
$$d(x(\lambda),x(\lambda_0)) = \lim_{n\to \infty}d(f_{\lambda}^n(p),f_{\lambda_0}^n(p)).$$
Essentially I have to prove the limits $\lim_{\lambda \to \lambda_0} $ and $\lim_{n\to \infty}$ commutes on the identity above, but I do not know how.
 A: Suppose for contradiction that the mapping $\lambda \mapsto x(\lambda)$ is not continuous at $\lambda_0$. Then there exists an $\epsilon > 0$ such that for all $\delta > 0$, there exists a $\lambda_\delta$ such that $$d(\lambda_\delta, \lambda_0) < \delta \  \ \  {\rm  but } \ \ \ d(x(\lambda_\delta), x(\lambda_0)) > \epsilon.$$
Now observe that
$$d\left( x(\lambda_\delta), F(x(\lambda_0), \lambda_\delta)\right) = d(F(x(\lambda_\delta), \lambda_\delta), F(x(\lambda_0), \lambda_\delta))\leq kd(x(\lambda_\delta), x(\lambda_0)),$$
by the contraction property.
Furthermore, we have
$$ d(F(x(\lambda_0), \lambda_\delta), F(x(\lambda_0), \lambda_0) =  d(F(x(\lambda_0), \lambda_\delta) , x(\lambda_0)), $$
and, by the triangle inequality, we obtain,
\begin{multline} d(F(x(\lambda_0), \lambda_\delta) , x(\lambda_0))\geq d(x(\lambda_\delta), x(\lambda_0)) -  d( x(\lambda_\delta) , F(x(\lambda_0), \lambda_\delta))\\  \geq (1-k)d(x(\lambda_\delta), x(\lambda_0)) \geq(1-k)\epsilon > 0.\end{multline}
Thus for every $\delta > 0$, we have $$d((x(\lambda_0), \lambda_\delta), (x(\lambda_0), \lambda_0)) = d(\lambda_\delta, \lambda_0) < \delta$$ but $$d(F(x(\lambda_0), \lambda_\delta), F(x(\lambda_0), \lambda_0) ) \geq(1-k)\epsilon.$$
This contradicts the assumption that $F$ is continuous at $(x(\lambda_0), \lambda_0)$.
