If $∀ϵ>0:∃δ>0:|x−a|δ<⟹|f(x)−F(a)|<E$
this implies continuity, we always say that for all $E$ we can find a delta. But for all delta we can find a Epsilon right? (this seems okay for me,even i'm not finding such affirmation) if is true what I said, given an interval $[a,b]$ catch $*xo*= \frac{a+b}{2}$ we have now $a------(a+b)/2-------b$ now since $*xo*$ E $[a,b]$ the definition of continuity fits,so if I choose delta= a-xo, i would stay in the closed interval and in all points this would be bounded for a $E$.
I bet this is wrong, but where i can find.
This is the weak step of your proof. It is not generally true that $$\forall x \exists y : P(x, y) \implies \forall y \exists x : P(x, y)$$
For a specific counterexample regarding continuity, consider the function $f(x) = \frac{1}{x}$. This function is continuous at $a = 1$, and yet for $\delta = 1$ there is no $\varepsilon$ such that $|x-1| < 1 \implies |f(x)-1| < \varepsilon$
(Note that in this counterexample the value of $\delta$ was specifically chosen to exploit a discontinuity of $f$ at $x = 0$. If you restrict the values of $\delta$ to ones where $f$ is continuous at the interval $[a-\delta, a+\delta]$, then it is true that for all $\delta$ you can find an $\varepsilon$, as it is indeed a theorem that continuous functions on closed intervals are bounded. But you can't use this fact while proving the theorem, as it would be circular reasoning)