Let me think about this a bit in abstract terms.
If the limit for x going to a of f(x) = f(a) for every x in some interval I, we say that f is continuous on I. The definition of limit then says that: for every a in I, and for every epsilon, we can find a delta (depending on epsilon and a), such that ... "
Uniform continuity is a bit stronger. It says that for all points a we can choose the same delta. So the statement is "for every epsilon, we can find a delta (depending on epsilon), such that ... for all a in I simultaneously".
In general, continuity does not imply uniform continuity. You cannot simply take the minimum of all the deltas in your interval, for example, because in general this does not exist. Technically, you should be talking about an infimum, but this might be zero and that is not allowed.
Now you find yourself in the situation, where you know that for some infinite interval there is a strictly positive delta which works for all the points in that interval. This suggests to me a proof in two steps:
* find a strictly positive delta on the finite closed interval [0, a] (basically proving that any continuous function on [a, b] is uniformly continuous)
* combine these two deltas into a single one which is > 0 and works for all of [0, ∞)