Uniformly continuous, continued

Aha!  Class on Friday revealed some important information in how this proof works out.   

Since $D$ is a bounded set, there are $x_1, x_2, ... , x_n \in D$ such that
$D\subset \cup^n _{i=1} (x_i - \delta, x_i + \delta)$

Because D is bounded, we can find an open cover: a collection of open sets that contains the original set.

For all $x \in D$, then $x_i - \delta \leq x \leq x_i + \delta$
Then $|x - x_i| < \delta$
Which implies $|f(x) - f(x_i)| < 1$
     But that just means $f(x_i) - 1 \leq f(x) \leq f(x_i) + 1$

Therefore, $f(D)$ is bounded.