I think you will see the difference if we explicitly write out the definitions for continuity and uniform continuity.
Let $f: X \rightarrow \mathbb{R}$ be a function from some subset $X$ of $\mathbb{R}$ to $\mathbb{R}$.
Def 1: We say that $f$ is continuous at $x$ if for every $\epsilon > 0$ there exists $\delta > 0$ such for every $y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
Def 2: We say that $f$ is continuous if $f$ is continuous at every $x \in X$.
This definition corresponds to what you called "continuity on a set" in your question.
Def 3: We say that $f$ is uniformly continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that for any two points $x, y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
As you have already noted in your question, the difference between uniform continuity and ordinary continuity is that in uniform continuity $\delta$ must depend only on $\epsilon$, whereas in ordinary continuity, $\delta$ can depend on both $x$ and $\epsilon$.
To illustrate this with a concrete example, let's say $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$, and $f(10) = 15, f(20) = 95$ and $f(30) = 10$.
In uniform continuity, if I pick $\epsilon > 0$, you must give me a single $\delta > 0$ such that the following statements all hold:
- If $|y - 10| < \delta$ then $|f(y) - 15| < \epsilon$.
- If $|y - 20| < \delta$ then $|f(y) - 95| < \epsilon$.
- If $|y - 30| < \delta$ then $|f(y) - 10| < \epsilon$.
On the other hand, in ordinary continuity, given the same $\epsilon > 0$ as above, you are free to pick three different deltas, call them $\delta_1, \delta_2, \delta_3$ such that the following statements all hold:
- If $|y - 10| < \delta_1$ then $|f(y) - 15| < \epsilon$.
- If $|y - 20| < \delta_2$ then $|f(y) - 95| < \epsilon$.
- If $|y - 30| < \delta_3$ then $|f(y) - 10| < \epsilon$.
To answer the second part of your question, as Jacob has already pointed out, your statement is the sequential characterisation of ordinary continuity. Since ordinary continuity does not imply uniform continuity the "if" part of your statement is false. However, it is easy to see that uniform continuity implies ordinary continuity, so the "only if" part of your statement is true.
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.
Best Answer
Let's think about what the two statements are saying in simple terms:
This means that for every $\epsilon$, there is an $\delta_\epsilon$ such that whenever the second condition is satisfied, that is whenever we have points $x$ such that $|x - c| < \delta_\epsilon$ holds, then we have $|f(x) - f(c)| < \epsilon$. So this is a statement about what we can say about $f(x)$ with regards to $f(c)$ whenever points $x$ satisfies a particular property.
What this second part is showing is that the property that we wants holds for certain $x_n$'s when $n \geq n_\epsilon$. Basically this is showing that the $x_n$'s have the property we desire, and thus the conclusion of the first statement holds.