Conclusion 2) in OP is incorrect, and everything past 1), and the sentence previous to 1), should be discarded.

Revised 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→∞, implies uniform continuity.

A proof of 2) is given by slipeternal in post #19 of

Uniform continuity

A clarified and expanded version is given below. If the expansion is part of the standard proof, then my apologies for posting it Peer Math Review, and this post will just be a clean-up of the previous one.

1) Let b be smallest x st |f(x) - k│ < ε/4, x>b. Then

|f(x) - f(y)│ < ε/2 for all │x - y│in [b,∞). Because lim f(x)=k as x →∞.

2) There is a δ such that |f(x) - f(y)│ < ε/2 for │x - y│< δ, x,y in[a,b]. Heine-Borel.

3) If x<b and y>b, |f(x) - f(y)│≤ |f(x) - f(b)│+ |f(y) - f(b)│< ε/2+ ε/2 = ε for │x - b │< δ and all y>b, or │x - y│< δ because │x - y│< δ → │x - b │< δ.

4) Suppose b’>b. Then (abbreviated)

a) |f(x) - f(y)│ < ε/2 for all │x - y│in [b’,∞)

b) |f(x) - f(y)│< ε/2 for all │x - y│in [b,b’].

c) |f(x) - f(y)│< ε/2 for │x - y│< δ all x,y in [a,b].

So

|f(x) - f(y)│< ε for │x - y│< δ for all x,y in [b,∞).

The last step is very important because it shows δ depends on a single b.

So continuity implies uniform continuity is consistent for the both conditions, on [a,b], and on [a,∞) with limf(x)=k.