= Solution
The <contraction mapping theorem> states that a <contraction mapping> $f$ of a nonempty <complete metric space> into itself has a unique <fixed point>. More explicitly, if $d(f(x),f(y))\leq qd(x,y)$ with $0\leq q<1$, iteration from any $x_0$ converges to that <fixed point>.
To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\leq q^nd(x_1,x_0)$, so for $m>n$ the <triangle inequality> gives
$$
d(x_m,x_n)\leq\sum_{j=n}^{m-1}q^jd(x_1,x_0)\leq\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus $(x_n)$ is a <Cauchy sequence>, and <completeness> supplies a limit $x_*$. A <contraction mapping> is <continuous>, so $f(x_*)=\lim f(x_n)=\lim x_{n+1}=x_*$. Two <fixed points> satisfy $d(x_*,y_*)\leq qd(x_*,y_*)$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_*)\leq q^nd(x_1,x_0)/(1-q)$.
For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the <uniform metric> $\rho(f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho(f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\leq d(f(x),g(x))+d(g(x),h(x))$ proves the <triangle inequality>. Hence $\rho$ is a <metric>.
Suppose $(f_n)$ is a <Cauchy sequence> for this <uniform metric>. At each $x$, $d(f_n(x),f_m(x))\leq\rho(f_n,f_m)$, so <completeness> of $X$ supplies a value $f(x)=\lim_n f_n(x)$. Given $\varepsilon>0$, choose $N$ with $\rho(f_n,f_m)<\varepsilon/2$ for all $n,m\geq N$. Fix $n\geq N$ and let $m\to\infty$ pointwise; then $d(f_n(x),f(x))\leq\varepsilon/2$ for every $x$. Thus $\rho(f_n,f)\leq\varepsilon/2<\varepsilon$. The <uniform limit theorem> makes this limit continuous, so $f\in F$. This proves the <completeness of the continuous-map space in the uniform metric>.
Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta(f)$ and $x_g=\theta(g)$. Using the <fixed point> equations,
$$
d(x_g,x_f)\leq d(g(x_g),f(x_g))+d(f(x_g),f(x_f))\leq\rho(f,g)+q_fd(x_g,x_f).
$$
Therefore the <fixed-point stability in the uniform metric> estimate is
$$
\boxed{d(\theta(g),\theta(f))\leq\frac{\rho(f,g)}{1-q_f}.}
$$
Choosing $\rho(f,g)<(1-q_f)\varepsilon$ proves \b[continuity of $\theta$ at every $f$]. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.
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