Q:

f is a real valued uniformly continuous function defined over a bounded subset A of R. Prove that the range of f is bounded in R.

If A is closed, it is compact. Therefore the range is compact. Since the range is contained in R, it is bounded.
Assume A is not closed and f is unbounded over A. Therefore there exists a sequence a(n) in A such that  f(a(n)) tends to infinity(or negative infinity). I assume the case f(a(n)) tends to infinity.
a(n) has a convergent subsequence b(n) with limit b. f(b(n)) tends to infinity. Therefore b does not belong to A. Since f(b(n)) will tend to f(b) if b belongs to A. 
I am not able to proceed beyond this point. I am looking for an explaination which uses the sequential criteria for uniformly continuous functions.
Any help will be appreciated. Thanks.

A:

Since $f$ is uniformly continuous, it maps Cauchy sequences to Cauchy sequences.  Therefore we can extend $f$ to a continuous map on the closure of $A$, which is a compact set, and its image under this extension is also compact.

