Show that the upper envelope of diff EQ solutions is a solution 
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.
 A: 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,  $\eta'(x)=f(x,\eta(x))$.
