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.
First, uniform continuity is about an interval, not just a point. You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$. Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement: for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.
Now what they never tell you is that $\delta$ depends on $x$. For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$. It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper. That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller, to get $|f(x) -f(c)| < \varepsilon$.
What uniform continuity means is that the $\delta$ does NOT depend on x. You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.
Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$. Steepness doesn't only mean that $f$ may go off to infinity in your interval. It could also oscillate around in some unpleasant way. Look for example at $f(x) = \sin(1/x)$ on $(0,1)$.
Best Answer
If in your proof the value of $\delta$ doesn't depend on your choice of $x$, but uniquely on $\varepsilon$, then you could have fixed $x$ after fixing $\varepsilon$ and determining $\delta$, therefore proving absolute continuity of the function, hence you can basically use the same proof, except for fixing $x$ and $y$ after determining $\delta$.