Assume the IVP $$ y'=\sqrt{x-y}=f(x,y), \quad y(2)=2 $$ Since there's no region of the form $$ R=\Big\{(x,y): |x-2|\leq 2 , |y-2| \leq 2 \Big\} $$ in which $f$ is continuous, does it mean that Picard's theorem is undecidable about existence?
Could another theorem tell us for sure?
On
Introduce $u=x-y$, then your equation leads to $$ u'=1-\sqrt{|u|} $$ which is separable. As the right side is different from $0$ close to $u(2)=0$, the integral in $$ G(u)=\int_0^u \frac{dv}{1-\sqrt{|v|}}=t-2 $$ exists and the anti-derivative can be locally and uniquely solved for $u$ by the inverse function theorem, as $G'(0)=1$.
Consider instead the equation $y' = \sqrt{|x-y|}$, where the right side is continuous on $\mathbb R \times \mathbb R$. Thus Picard says the initial value problem $y' = \sqrt{|x-y|}, y(2)=2$ has a solution in some interval $(2-\delta, 2+\delta)$. It's not hard to show that this solution satisfies $y \le x$ for $x \in [2, 2+\delta)$, therefore it is a solution to your initial value problem in that interval. On the other hand, for $x$ slightly less than $2$ we would have $y > x$, so it can't satisfy your differential equation there.