Prove that uniformly continuous function preserves boundedness of an interval Assume that a function $f$ is uniformly continuous on $(a,b)$. Prove that 
a)there is a $\delta$ such that for all $x_0$, for all $x \in (a,b)\cap(x_0-\delta,x_0+\delta)$,$|f(x)| \leq |f(x_0)|+1$
b)$f(x)$ is bounded on $(a,b)$
I manage to prove part (a) but not part (b). Can anyone guide me ?
EDIT: For part b), can I prove like this? For $\forall x \in (a,b)$, take $\delta$ from part a) such that $$(a,b) \subset \bigcup_{i=1}^{n}{V_\delta(x)}$$ Then $y \in (a,b)$ is contained in at least one of the delta neighbourhood of $x$. By part a), we have $|f(y)| \leq |f(x)|+1$. Take $M=\max_{1 \leq i \leq n}{\lbrace x_i \rbrace}$. Then $f$ is bounded for all $y \in (a,b)$. 
One thing I would like to ask here is that how do we know that the union of delta neighbourhood is finite ? We know that there are infinitely many points in an open interval. 
 A: Method one: Note that $(a,b)$ is dense in $[a,b]$, and that any uniformly continuous function on a dense subset of a space can be extended to a uniformly continuous function of the larger space. Thus we can extend $f$ to a uniformly continuous function $g$ on $[a,b]$, which is compact. Hence the image of $g$ is compact, so bounded, so the image of $f$ is bounded as well.
Method two: Let $\delta$ be such that $|x-y|<\delta\implies |f(x)-f(y)|<1$. Let $c=\min(a+\delta,b)$. Clearly if $x\in (a,c]$ then $|f(x)-f(c)|<1$. Use the triangle inequality and induction to show that if $x\in [c,c+n\delta)$ then $|f(x)-f(c)|< n$. Thus $|f(x)-f(c)|<\max(1,\lceil (b-a-1)/\delta\rceil)$ for all $x\in (a,b)$, so
$$f(c)-\max(1,\lceil (b-a-1)/\delta\rceil)< f(x)<f(c)+\max(1,\lceil (b-a-1)/\delta\rceil).$$
Method three (using part a): Let $\delta$ be as in part a and $x_0=\frac{a+b}{2}$. You have already shown that
$$x\in (x_0-\delta,x_0+\delta)\implies |f(x)-f(x_0)|<1.$$
Suppose that 
$$x\in (x_0-n\delta,x_0+n\delta)\implies |f(x)-f(x_0)|<n,$$
and let $x\in (x_0-(n+1)\delta,x_0+(n+1)\delta)$. We want to show that $|f(x)-f(x_0)|<n+1$. If $x<x_0$, then certainly $x+\delta\in (x_0-n\delta,x_0+n\delta)$ so 
$$|f(x)-f(x_0)|\leq |f(x)-f(x+\delta)|+|f(x+\delta)-f(x_0)|<1+n$$
as desired. Note that $|f(x)-f(x+\delta)|<1$ is using $x$ as the $x_0$ in part $a$. I'll leave the case $x>x_0$ to you. By induction this shows that
$$x\in (x_0-n\delta,x_0+n\delta)\implies |f(x)-f(x_0)|<n$$
for all $n$. Thus if we let $N=\lceil\frac{b-a}{2}/\delta\rceil$ we get 
$$x\in (a,b)\implies x\in (x_0-N\delta,x_0+N\delta)\implies |f(x)-f(x_0)|<N$$
thus we have $f(x_0)-N<f(x)<f(x_0)+N$.
