Let S be the set of points on an interval and let f be continuous on S. Then at every point x there is a neighborhood of x st

|f(x)-f(y)|<ε if |x-y|<δ(x).

The set of all δ(x) form an open cover of S. If S is compact, (closed and bounded, [a,b]), then there is a finite collection of the δ(x) which covers S. Then δ = min δi(x) and f is uniformly continuous. That establishes sufficiency. Necesssity established by uniform continuity implies continuity.

If S is not bounded, we have to choose a minimum from an infinite collection of δ(x). If you show that δ(x) < δ for all x, you haven’t determined that a min δ exists for all x and you haven’t proved uniform continuity.*

Conclusion:

1) f(x) continuous on S compact ([a,b]) iff f(x) uniformly continuous.

2) f(x) continuous on [a,∞) and lim f(x)=k as x→∞, does not imply uniform continuity.

*Given all the δ(x) < δ, is there a minimum one(s) among them- a subtle point. For example, if you can show for a given ε, 0 < δ(x) < α, what is the minimum value of δ(x)?

EDIT: There is an ambiguity in the definition of Uniform Continuity, depending on whether, given ε for the interval, you require:

3) A SPECIFIC MIN δ(ε), or

4) ANY δ LESS THAN δ(ε)

The conclusions 1) and 2) above assume definition 3).