I wonder what if we could change the order of the quantifiers in the definition of continuity of function. I mean
$a)$ For any number $\forall \delta >0,$ there exists some number $ \exists \varepsilon =\varepsilon \left( \delta ,x_{ 0 } \right) \quad$ such that $\quad \left| x-{ x }_{ 0 } \right| <\delta \quad \Rightarrow \left| f\left( x \right)-f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad$
$ b)$ For any number $ \forall \varepsilon >0, $ there exists some number $\exists \delta =\delta \left( \varepsilon ,x_{ 0 } \right) \quad $ such that $\quad \left| f\left( x \right) -f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad \Rightarrow \quad \left| x-{ x }_{ 0 } \right| <\delta \quad $
$c)$ For any number $ \forall \delta >0, $ there exists some number $\exists \varepsilon =\varepsilon \left( \delta ,x_{ 0 } \right) \quad $ such that $\quad \left| f\left( x \right) -f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad \quad \Rightarrow \quad \left| x-{ x }_{ 0 } \right| <\delta $
I know in these variants function is not continuous at $x=x_0$ point, but I can't prove it, or can't get a really good counterexample.
a) This is satisfied if e.g. function $f$ is bounded.
b) Let $f(x)=g(x)x$ where $g$ is any function that takes values in $\mathbb R\setminus[-1,1]$.
Then $|f(x)-f(x_0)|\geq |x-x_0|$ so that $\delta=\varepsilon$ works.
c) Let $f$ be a function that satisfies $x\neq x_0\implies |f(x)-f(x_0)|>1$.
Then $\varepsilon=1$ works.
In all cases $f$ can be chosen to be not continuous at $x_0$.