Let $(X,d)$ be a complete metric space, and let $f: X \to X$ and $g : X \to X$ be two strict contractions on $X$ with contraction coefficients $c$ and $c'$ respectively. Then, by the contraction mapping theorem, $f$ has some fixed point $x_0$, and $g$ has some fixed point $y_0$. Suppose that we know that there is an $\epsilon>0$ such that $d(f(x), g(x)) \le \epsilon$ for all $x \in X$ (i.e., $f$ and $g$ are within $\epsilon$ of each other in the uniform metric). Show that $d(x_0, y_0) \le \epsilon/(1- \min(c,c'))$. Thus nearby contractions have nearby fixed points.
$d(x_0, y_0) = d(f(x_0), g(y_0)) \le d(f(x_0), f(x)) + d(f(x), g(x)) + d(g(x), g(y_0)) \le cd(x_0, x) +\epsilon + c'd(x, y_0) $.
Therefore, I need to show that $cd(x_0, x) +\epsilon + c'd(x, y_0) \le \epsilon + \min(c, c')d(x_0, y_0)$. I think that this inequality is not true because $\min(c, c')d(x_0, x) + \min(c, c')d(x, y_0)\le cd(x_0, x) +c'd(x, y_0)$, and $d(x_0, y_0) \le d(x_0, x) + d(x, y_0)$.
I would appreciate if you give some help.
Define inductively $f^0(x)=x$ and $f^{n+1}(x)=(f \circ f^n)(x)$.
Then we have
$$d(f^n(x), f^n(y)) \le c^n d(x, y)$$
and also
\begin{equation} \label{eq1} \begin{split} d(f^n(x), x) & \le \sum_{k=0}^{n-1} d(f^{k+1}(x), f^k(x)) \\ & \le \sum_{k=0}^{n-1} c^k d(f(x), x) \\ & \le \frac{ d(f(x), x) }{ 1 - c }. \end{split} \end{equation}
Hence
\begin{equation} \label{eq2} \begin{split} d(x_0, y_0) & = d(f(x_0), y_0) \\ & = d(f^n(x_0), y_0) \\ & \le d(f^n(x_0), f^n(y_0)) + d(f^n(y_0), y_0) \\ & \le c^n d(x_0, y_0) + \frac{ d(f(y_0), y_0) }{ 1 - c } \\ & = c^n d(x_0, y_0) + \frac{ d(f(y_0), g(y_0)) }{ 1 - c } \\ & \le c^n d(x_0, y_0) + \frac{ \epsilon }{ 1 - c }. \end{split} \end{equation}
Letting $n \to \infty$, we get $d(x_0, y_0) \le \frac{ \epsilon }{ 1 - c }$. Analogously, we find $d(x_0, y_0) \le \frac{ \epsilon }{ 1 - c' }$.
Therefore, $d(x_0, y_0) \le \frac{ \epsilon }{ \max(1 - c, 1 - c') } = \frac{ \epsilon }{ 1 - \min(c, c') }$.