Uniformly continuous

I will begin with a definition of uniformly continuous directly from the book

A function $f: D \to R$ is uniformly continuous on $E\subset D$ iff for every $\epsilon >0$, there is \delta >0 such that if $x,y\in D$ with $|x-y|<\delta$, then $|f(x) - f(y)|<\epsilon$. If $f$ is uniformly continuous on $D$, we say $f$ is uniformly continuous.

That definition is helpful in proving this theorem

3.9 If $f: D\to R$ is uniformly continuous and $D$ is a bounded set, then $f(D)$ is a bounded set.

Choose \epsilon =1. 
     Then \exists \delta >0 such that
          if $|x-y| <\delta$
          then $|f(x) - f(y)| <1$

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

Now we have $|x-x_i| <\delta$ where $x \in D$
     Then $|f(x) - f(x_i)| <1$

Thus, $f(D)$ is bounded!

Well, I find this proof easy to follow, except for the union of the neighborhoods.  How does that come from D being bounded?