continuity of a homotopy Let $X$ be a contractible compact Hausdorff space, i.e. for a fixed $x_0\in X$ there is a continuous map $\alpha:X\times [0,1]\to X $ with $\alpha (x,0)=x$ and $\alpha(x,1)=x_0$ for all $x\in X$. 
Consider the $C^*$-algebras $C(X)$ and $\mathbb{C}$. My aim is to prove that $C(X)$ is homotopy equivalent to $\mathbb{C}$. 
For this consider the $*$-homomorphisms $ev_{x_0}:C(X)\to\mathbb{C},\;  f\mapsto f(x_0)$ and $\psi:\mathbb{C}\to C(X), \lambda \mapsto \lambda 1_{C(X)}$ (with $1_{C(X)}(x):=1_\mathbb{C}$ for all $x\in X$). We have $ev_{x_0}\circ \psi=id_{\mathbb{C}}$, so that we only need a homotopy between $\psi\circ ev_{x_0}$ and $id_{C(X)}$. The homotopy between $\psi\circ ev_{x_0}$ and $id_{C(X)}$ is defined as follows:
$$H:C(X)\times [0,1]\to C(X),\; H(f,t)=f(\alpha(\cdot ,t)).$$
Indeed, it is $F(f,0)=f(\alpha(\cdot ,0))=f(\cdot)$ and $F(f,1)=f(\alpha(\cdot ,1))=f(x_0)1_{C(X)}$.  But I still have to proof that $$H_{\cdot}(f):[0,1]\to C(X),\;\; t\mapsto H(f,t)=f(\alpha(\cdot ,t))$$ is continuous for all $f\in C(X)$.
For this, let $\epsilon >0$ and $t_0\in [0,1]$ fixed but arbitrary. Claim: There exists a $\delta >0$ such that for all $t\in [0,1]$ with $|t-t_0|<\delta$ it follows $\|H(f,t)-H(f,t_0)\|_{\infty}=\sup\limits_{x\in X}\|f(\alpha(x ,t))-f(\alpha(x ,t_0))|<\epsilon$.
I don't know how to choose $\delta$ correctly. My attempt is to use that $\alpha_x:[0,1]\to X$, $\alpha_x(t)=\alpha(x,t)$ is (uniformly) continuous for all $x\in X$. Then for every neighborhood of $\alpha_x(t_0)$, $U(\alpha_x(t_0))$, there exists a $\delta_1 >0$ such that for all $t\in [0,1]$ with $|t-t_0|<\delta_1$ it follows $\alpha_x(t)\in U (\alpha_x(t_0))$. Now I don't know how to continue. Shall I use the continuity of $f\in C(X)$? Or how to prve the continuity of $H_{\cdot}(f)$ for all $f\in C(X)$?
 A: Let $g:=f\circ\alpha : X\times[0,1]\to\mathbb{C}$. First consider the cover of $X\times[0,1]$ with the open sets $g^{-1}(B_{\epsilon/2}(z))$ for $z\in\mathbb{C}$ arbitrary. Now every $(x,t)\in X\times[0,1]$ is contained in one of these sets and therefore there exists an open neighbourhood $U$ of $x$ and a $\delta>0$ such that $g(U\times B_\delta(t)) \subseteq B_{\epsilon/2}(z)$ for some $z\in\mathbb{C}$. Then the sets $U\times B_{\delta/2}(t)$ are still a covering of $X\times[0,1]$. Choose finitely many of these products to cover $X$ with, i.e.
$$X=\bigcup_{i=1}^m U_i \times B_{\delta_i/2}(t_i)$$
and set $\delta:=\min_{i} \delta_i/2$. I claim that this $\delta$ is what you want: If $x\in X$ is arbitrary and $|t-t'|<\delta$, then there exists some $i$ with $(x,t)\in U_i\times B_{\delta_i/2}(t_i)\subseteq U_i\times B_{\delta_i}(t_i)$. Because $|t-t'|<\delta_i/2$ we also have $(x,t')\in U_i\times B_{\delta_i}(t_i)$. By construction $g(x,t)$ and $g(x,t')$ are both contained in $B_{\epsilon/2}(z)$ and thus have distance less than $\epsilon$ from each other.
