Is the mapping sending the parameter to the corresponding fixed point continuous? 
Let $(X,d)$ be a complete metric space, let $T$ be a topological space, and $(f_t)_{t\in T}$ be a family of mappings $f_t:X\to X$ with the following properties : for each $x\in X$, the map $T\to X,t\mapsto f_t(x)$ is continuous, and there exists a constant $k$ such that
$$0<k<1,\;d(f_t(x),f_t(y))\le kd(x,y)\;\;\forall x,y\in X,\forall t\in T. $$

By the Banach fixed point theorem, for each $t\in T$ there exists a unique fixed point $x_t$ of $f_t$. Is the map $T\to X,t\mapsto x_t$ continuous ?
I think that if $T$ is a metric space then the answer is true. Now, suppose $T$ is just a topological space, is the mapping still continuous ?
 A: I have a partial answer, with a constraint that seems quite reasonable.
Let's define $f:T\times X \rightarrow X$ so that $f(t, x)=f_t(x)$ and let's assume that $f$ is continuous with respect to the product topology (this is the constraint).
To fix notation, let's call $f^*:T\rightarrow X$ the function that maps each $t\in T$ to the fixed point $f^*(t)\in X$ of $f_t:=f(t, -):X\rightarrow X$. Now, by Banach fixed point theorem we know that $f^*(t)=\lim_{n\to\infty} f_t^n(x')$, where $x'$ is any point of $X$.
Now, I'm going to prove that $f^*$ is continuous by showing that, chosen any open set $V\subseteq X$, we can always find an open $U\subseteq T$ such that $f^*(U)\subseteq V$.
First, let's fix $t_0\in T$, $x_0\in X$ such that $f^*(t_0)=x_0$ and $\varepsilon > 0$. Let $B(x_0, \varepsilon)$ the ball centered in $x_0$ with radius $\varepsilon$. Since $f$ is continuous, we know that $f^{-1}(B(x_0,\varepsilon))\subseteq T\times X$ is open. By definition of product topology, since $(t_0, x_0)\in f^{-1}(x_0)$, we can find two open sets $U\subseteq T$ and $W\subseteq X$ such that $(t_0, x_0)\in U\times W\subseteq f^{-1}(B(x_0,\varepsilon))$. Since both $W$ and $B(x_0,\varepsilon)$ contains $x_0$, we can (finally!) assume that $W\subseteq B(x_0,\varepsilon)$ (taking the intersection if necessary).
Now, let $t\in U$. Using triangular inequality, it's easy to prove by induction that
$$
d(f^n_t(x_0),x_0)\leq\varepsilon(1+k+k^2+...+k^{n-1})=\varepsilon\frac{1-k^n}{1-k}<\frac{\varepsilon}{1-k}.
$$
For example, if $n=2$ we have
$$
d(f^2_t(x_0),x_0)\leq d(f^2_t(x_0),f_t(x_0)) + d(f_t(x_0),x_0) \leq kd(f_t(x_0),x_0) + d(f_t(x_0),x_0) = \varepsilon(1+k),
$$
where we used triangular inequality for the first inequality, the Lipschitz condition for the second one and the fact that $f_t(x_0)\in B(x_0,\varepsilon)$ for the last equality.
So for all $n\geq 1$ we have that $d(f_t^n(x_0),x_0)<\frac{\varepsilon}{1-k}$.
Hence
$$
d(f^*(t),f^*(t_0)) \leq \frac{\varepsilon}{1-k}.
$$
In particular, we then have that $f^*(U)\subseteq B(f^*(t_0), \frac{\varepsilon}{1-k})$. Since we chose $\varepsilon$ arbitrarily, we can always choose it small enough so that $B(f^*(t_0), \frac{\varepsilon}{1-k})\subseteq V$, which is your thesis.
I wasn't able to find a proof without assuming $f$ to be continuous as I did and I highly suspect that it is not true in that generality.
