Update
Since I wrote the original post, I have received a lot of answers. However, all of them say why my alternatives do not work. None of them directly answer the question I asked. What makes the definition of continuity superior to the others? I am more interested in knowing why the epsilon-delta definition of continuity works rather than knowing why the variants do not work.
Original
I want to expand on a question that was asked a few years ago.
The $\epsilon, \delta$-definition of continuity is:
$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$
What makes the definiton above superior to all of the ones below?
$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta $$
$$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$
$$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta $$
And, finally, the one that nags me the most (considering most "intuitive" explanations I have heard, like the one at Khan Academy, this should be almost equivalent to the one above): $$\forall \delta > 0\ \exists \varepsilon > 0\ \text{s.t. } 0< |f(x) - f(x_0)| < \varepsilon \implies |x - x_0| < \delta $$
I can go on making other variations, but I reckon you get the point...