Let $(M,d)$ a metric space, a function $f:M\to M$, is a contraction if there exists $k\in [0,1)$, such that $d(f(x),f(y))\leq k\,d(x,y)$. If $(M,d)$ is complete, one can show that $f$ has a unique fixed point (the Banach fixed point theorem).
Now consider $(f_n)$, a sequence of contractions on $(M,d)$, a complete space, such that $(f_n)\to f$ uniformly (i.e. with respect to the metric $d'(f,g)=\sup\{d(f(x),g(x))\mid x,y\in M\}$). I want to show that the sequence $(x_n)$ of the fixed points of the $f_n$ converges to a fixed point of $f$; I need show that $(x_n)$ is a Cauchy sequence, with that, we have that $x_n\to p$, for some $p$, also $f(x_n)\to f(p)$, and $f(x_n)\to p$, because $d(f(x_n),p)\leq d(f(x_n),x_n)+d(x_n,p)=d(f(x_n),f_n(x_n))+d(x_n,p)$, so $f(p)=p$.
I appreciate any suggestion on how to show that $(x_n)$ is a Cauchy sequence, thanks.
Let $k_m$ be the contraction constant of $f_m$. Then \begin{align} d(x_m,x_n)&=d(f_m(x_m),f_n(x_n)) \\ &\leq d(f_m(x_m),f_m(x_n))+d(f_m(x_n),f_n(x_n)) \\ &\leq k_md(x_m,x_n)+d'(f_m,f_n) \end{align} for all $m,n\in\mathbb{N}$. Since $k_m<1$, we may rearrange to get \begin{align} d(x_m,x_n)\leq\frac{d'(f_m,f_n)}{1-k_m}. \end{align} We are done if $\limsup k_m<1$. Otherwise counterexamples exist: Consider contraction mappings on $[0,1]$ defined by \begin{align} f_n=\begin{cases} (1-1/n)x &\text{if $n$ is odd}, \\ (1-1/n)x+1/n &\text{if $n$ is even}. \end{cases} \end{align} Then $x_n=0$ if $n$ is odd and $x_n=1$ if $n$ is even, so $(x_n)$ does not converge. However, $f_n$ converges uniformly to the identity function on $[0,1]$.