I don't understand uniform continuity :(

I don't understand why the last part is wrong :(. Where did 'i' come to play?
This is what I've done step by step written down in propositional calculus:

Suppose that [itex]\forall x_n, \forall y_n:|x_n - y_n| \to 0 \implies |f(x_n)-f(y_n)| \to 0[/itex] is false:
1) Negation: [itex]\exists x_n, \exists y_n: |x_n - y_n|\to 0 \land \neg(|f(x_n)-f(y_n)| \to 0)[/itex]
2) conjunction elimination: [itex]\exists x_n, \exists y_n: \neg(|f(x_n)-f(y_n)| \to 0)[/itex]
3) [itex] \neg(|f(x_n)-f(y_n)| \to 0) \equiv \forall\delta>0, \exists\epsilon>0 : |x_n - y_n|<\delta \implies |f(x_n)-f(y_n)|\geq \epsilon[/itex]
4) [itex] \exists x_n, \exists y_n, \forall \delta>0, \exists \epsilon>0: |x_n-y_n|<\delta \implies |f(x_n)-f(y_n)| \geq \epsilon[/itex]
5) [itex] \exists x_n, \exists y_n, \forall \delta>0, \exists \epsilon>0: |x_n-y_n|<\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon>0, \exists \delta>0, \forall x_n, \forall y_n : |x_n - y_n|<\delta \implies |f(x_n)-f(y_n)|<\epsilon)[/itex]

This is what my argument was in symbolic logic, I usually suck at propositional calculus but when I write things in that way I find it easier to argue with people.

For a constant n it is impossible that |x_n - y_n| < delta for all delta(unless x_n = y_n).

I'm not sure where you screwed up but it's not right.

hmmm, sounds you're right. Maybe I should've put up a quantifier like for any n in natural numbers....
well, except that part, do I have other mistakes? How would you rephrase my proof to make it right?

Very great sentences.
Well, the last part is important, |x| is not differentiable everywhere (it's not differentiable at x=0) but yet it is uniformly continuous over R. Right? I used your intuitive method.

I think your first statement does not correctly state what you wanted it to state.

Yes, exactly.

If you can "see" without any need for explanation or symbols that 1/x and e^x are continuous but not uniformly continuous, that's what I'm trying to explain.

Because as x gets larger, e^x gets a LOT LOT LOT larger. And as x gets close to zero, 1/x gets a LOT LOT LOT larger. Both 1/x and e^x get stretched.
Once you see that you can intuitively see that they must fail to be uniformly continuous.

Once you can see that, the epsilons and deltas will be easer.

Yea. me too. But I don't know how to reword my proof. So if we exclude the details, have I got the strategy right in my proof, or my proof is totally wrong?

The problem starts with the forall statements, the forall needs to be applied to the sequences, not the elements of the sequences.

Ahh, that's right. I guess my proof is flawed even more than that, because I didn't use the fact that |xⁿ-yⁿ| goes to zero.
if [itex] |x_n - y_n| \to 0 [/itex] then [itex] \exists M\in\mathbb{N}>0,\forall\epsilon>0,\forall n\in\mathbb{N}>M: n\geq M \implies |f(x_n)-f(y_n)| < \epsilon[/itex]
maybe this is where that i could come to play, because this time when I negate this there must exist n=i that has the property you said. Am I right?

yess