= Solution
Use <Picard iteration> on $[a-h,a+h]$:
$$
u_0(x)=b,\qquad u_{n+1}(x)=b+\int_a^x f(t,u_n(t))\,dt.
$$
By <induction>, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The <Lipschitz condition> gives, for $n\ge0$,
$$
|u_{n+1}(x)-u_n(x)|\le\frac{MA^n|x-a|^{n+1}}{(n+1)!}.
$$
For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.
The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The <Lipschitz condition> also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit gives
$$
u(x)=b+\int_a^x f(t,u(t))\,dt.
$$
The <fundamental theorem of calculus> makes $u$ a continuously differentiable solution with $u(a)=b$.
For uniqueness, let $u,v$ be two solutions and put $D_0=\sup|u-v|$ on the interval. Repeated use of the <Lipschitz condition> in their integral equations gives
$$
|u(x)-v(x)|\le D_0\frac{(A|x-a|)^n}{n!}
$$
for every $n$. Letting $n\to\infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore \b[a unique solution exists on the full stated interval]. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.
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