I kindly request feedback regarding the following proof. I am not sure about the inequalities, because of the two different metrics. What I essentially want to know is if the inequality makes sense, where i switched from $d_X$ to $d_K$? Firstly, the task is as follows:
Let $K(X)$ denote the set of contractions, i.e., $$K(X) = \{f: X \rightarrow X \mid \exists \kappa_f \in [0,1): d(f(x), f(y)) \leq \kappa_f d(x,y) \quad \forall x,y \in X\}.$$ We equip this set with the metric $$d_K(f, g) = \sup_{x \in X} \{d_X(f(x), g(x))\}.$$ Let $\Phi: K(X) \rightarrow X$ be the mapping that assigns to each contraction of $X$ its unique fixed point. Determine if $\Phi$ is continuous.
Proof:
We will show $\Phi$ is contious.
Let $\varepsilon > 0$; $f, g \in K(X)$; choose $\delta = \varepsilon$, $\delta > 0$. Such that $d_K(f, g) < \delta. $Let $a_i = \Phi(i(x))$.
\begin{align*} d_X(\Phi(f(x)), \Phi(g(x))) &= d_X(a_f, a_g) \\ &\leq \sup_{x \in X} \{d_X(f(x), g(x))\} \\ &= d_K(f, g) < \delta = \varepsilon. \end{align*}