This is the proposition 9.9.15:
Let $X$ be a subset of R, and let $f:X\rightarrow $R be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.
I wanted to use the notation $B_{u}(v) := \{ x : d(v,x) < u\}$ and this is my answer:
First of all, for continuity we have:
$$\forall \epsilon > 0 \,\,\exists\delta>0 : x \in B_{\delta}(y) \Longrightarrow x \in B_{\epsilon}(f(y)) \,\,\forall x,y \in X$$
Let's suppose $f(E)$ is not bounded: $$\forall \epsilon > 0 \,\,\exists x_0 \in E : |f(x_0)| > \epsilon$$
However, as $x_0 \in X$, we have this: $$\forall \epsilon > 0 \,\,\exists \delta >0 : x \in B_{\delta}(x_0) \Longrightarrow f(x) \in B_{\epsilon}(f(x_0))\,\, \forall x$$
Note that $x_0 \in B_{\delta}(x_0)$ then $f(x_0) \in> B_{\epsilon}(f(x_0))$ but $|f(x_0)| > \epsilon$ implies $f(x) \notin B_{\epsilon}(f(x_0))$. Finally, by contradiction, we can conclude that E must be bounded.
My problem with my answer is that I'm not using the fact that E is bounded. I'm not an expert, so I'd be grateful if you could show me other errors in this reasoning.