Assume $f(x,y)$ is continuous in the rectangle $$R:=\{(x,y)\in \mathbb{R}^2: |x-x_0|\leq a,\,|y-y_0|\leq b\},$$ and let $S$ be the set of functions $y=y(x)$ which are continuous, satisfy $y'(x)=f(x,y)$, and $|y(x)-y_0|\leq b$ for $|x-x_0|\leq a$. Show that $$\eta(x) := \sup_{y\in S}y(x)$$ is in $S$.
Here and here are some answers posted by the grader/instructor, which I don't understand. In the first one, he points out correctly that solutions cannot cross, and so the solutions in $S$ are ordered. I don't understand how he concludes from that there is a uniformly converging sequence. In the second, the handwriting makes it hard to understand how he's indexing things.
At one point, I thought we might want to use Dini's theorem, somehow, if we can show that the limit is continuous, but that seems like begging the question.
The provided solution is sketchy at best; written by a graduate student, I guess? The elements of $S$ are not ordered in general: for example, both $y=x^3$ and $y=0$ satisfy $y'=3y^{2/3}$ on the interval $[-1,1]$. It is true that $S$ is a lattice: the maximum or minimum of two elements of $S$ again lies in $S$. This is easy to check from the definition of $y'(x)$.
My solution. Since $f$ is uniformly bounded, the family $S$ is equicontinuous. Furthermore, since $f$ is uniformly continuous, the family $S$ is uniformly differentiable. Namely, for every $\epsilon>0$ there is $\delta>0$ such that for all $y\in S$, all $x\in [x_0-a,x_0+a]$ and for all nonzero $h\in (-\delta,\delta)$ we have $$\left|\frac{y(x+h)-y(x)}{h} - f(x,y(x)) \right|\le \epsilon \tag{1}$$ (provided that $x+h\in [x_0-a,x_0+a]$, of course.) To prove (1), apply the mean value theorem to $y$ and estimate $|f(\xi,y(\xi))-f(x,y(x))|$ using the continuity of $y$ and $f$.
Fix $x$ and $h$ as in (1). Consider a sequence $u_n\in S$ such that $u_n(x)\to\eta(x)$, and a sequence $v_n\in S$ such that $v_n(x+h)\to \eta(x+h)$. Let $y_n=\max(u_n,v_n)$, as noted above, $y_n\in S$. Put $y_n$ into (1) and pass to the limit: $$\left|\frac{\eta(x+h)-\eta(x)}{h} - f(x,\eta(x)) \right|\le \epsilon \tag{2}$$ Since (2) holds for all $x,h$ as above, $\eta'(x)=f(x,\eta(x))$.