Continuity and sequential continuity

I think the forward direction is fine (even then you could probably polish it a bit better), but the backward direction is a little confusing.

I know now that $\forall \epsilon > 0, \exists n_\epsilon : \forall n \geq n_\epsilon \implies |x_n - x| < \epsilon$

So for a certain $\epsilon$:

No that isn't what you actually know. What you do know is that

$$\lim_{n \to \infty} f(x_n) = f(x)$$

and we want to prove $$\lim_{z \to x} f(z) = f(x).$$

This one says that whenever we are on a neighborhood of $x$, the images are tolerably the same. I think you were too focused on "matching symbols" to make the proof work. user117042 pretty much outlined the proof for you. I'll only comment on why he chose $\delta = 1/n$.

By choosing $\delta = 1/n$, we create a convergent sequence such that its image must also get reasonably close to its target, but that all hinges on continuity, which we assume is not true, hence the contradiction.

Also, I've noticed you aren't using the metrics in $X$ and $Y$ at all and you are simply using absolute values. You need to fix that afterwards too before you submit your answer.

I think your proof is bad written. For the direction (=>) we can do straightforwardly. For the reverse direction, I suggest to use contradiction argument: negating the definition of continuity yields that there exists $\epsilon>0$ such that for any $\delta(\epsilon)>0$ ..Then we only need to choose $\delta=1/n$ and it then appears a sequence to deal with! (sorry for my mistake, I've corrected it)