Question: I have trouble relating two forms of finding the volume of a cylinder. First of all you can imagine integrating the area of a disk along the height. Something like this, $V=\int\_{0}^{h} \pi r^2 dx = \pi r^2h$. But also you can think of it as integrating the area of a rectangle ($hr$) along the "angles", so the volume would be $V=\int\_{0}^{2\pi} hrd\theta=2\pi hr$. I know the last one is wrong. But I really don't see why. Hope you can help

Answer: 
Your proof is correct, although a little verbose. Often times it is nice to give an explicit bound $|f(x)| \leq |f(c)|+\epsilon$. I don't think it is necessary to start with $x \in (c-\delta,c+\delta)$ and then deduce that $|x-c|<\delta$, that is evident.
From the definition

$A\subset \mathbb{R}, c\in A$

$\forall \varepsilon>0,\exists \delta >0,\forall x\in A \quad |x-c|<\delta \implies |f(x)-f(c)|<\varepsilon$

Let $\varepsilon >0$, so according to the definition, there is $\delta >0$ such as $\forall x\in (c-\delta, c+\delta)$ this inequality ($|f(x)-f(c)|<\varepsilon$) is satisfied.

then

$|f(x)-f(c)|<\varepsilon \iff f(c)-\varepsilon<f(x)<f(c)+\varepsilon$

$\implies f$ is bounded