Proving f is uniformly continuous

Good afternoon
I'd like to know if my proof is OK, I'm not sure it doesn't get into a
kind of circular reasoning, a loop, something like a circular
reference in a spread***.

If f:[0, oo) -> R is continuous and lim ( x-> oo) f(x) exists in R,
then f is uniformly continuous on [0, oo).

My proof:

Since lim ( x-> oo) f(x) exists, if follows from Cauchy criterium
there's k >0 such that |f(x1) - f(x2| < eps for every x1 and x2 in
[k , oo). Since [0 ,k] is compact (closed and bounded), f is is
uniformly continuous on [0 , k], because it's continuous. Then,
there's d >0 such that |f(x1) - f(x2| < eps for every x1 and x2 in [0,
k] such that |x2 - x1| < d. But we can have x1 in [0 , k] and x2 in
[k, oo). If this is the case and, in addition, we have |x2 - x1| = x2
- x1 < d, then we must have k - x1 < d and x2 - k <d. This implies |
f(k) - f(x1)| < eps and |f(x2) - f(k)| < eps. By the triangle
inequality, |f(x2) - f(x1)| <= |f(k) - f(x1)| + |f(x2) - f(k)| < eps
+ eps = 2eps.

So, for every x1 and x2 in [0, oo) satisfying |x2 - x1| < d, we have |
f(x2) - f(x1)| < 2eps. Since eps is arbitrary, f is uniformly
continuous on [0, oo). Well, it'd be more elegant if I had started my
proof with eps/2, to finish up with eps.

My question is: in my proof k depends on eps and d depends on eps and
k. So, we might say d depends only on eps, but to a same eps there can
correspond a lot of k's, actually infinitely many k's, its not a one-
to-one function. The number k can slide along the horizontal axis. I
made a choice (k) and then another choice (d) based on the first one.
Isn't this a sort of circularity that can invalidate my proof?

Thank you
Sharon

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