Q:

Prove that $f(x) = 4 - 15x$ is continuous

Need help showing this. 
Show that the function $f\colon \mathbb{R} \to \mathbb{R}:$ defined by  $f(x) = 4 - 15x$ is continuous at every point $c\in\mathbb{R}$. 
Thanks.

A:

Below I try to describe one way of thinking about this. The actual proof is written in blue, the rest is just guidance. I did it with a lot of detail, to the point where it might actually confuse you, so beware.

You are being asked to prove that
$$\forall c\in \mathbb R\forall \varepsilon>0\,\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right).$$
(This is just the definition, there is nothing to it).
(In words: given a real number $c$ and a positive real $\varepsilon$, there exists a positive real number $\delta$ with the property that for all real numbers $x$ such that $|x-c|<\delta$, it holds that $|(4-15x)-(4-15c)|<\varepsilon$).
(Since this is a universal statement, you begin by taking an arbitrary $c\in \mathbb R$.)
$\color{blue}{\text{Let $c\in \mathbb R$.}}$
(Now you wish to prove that $\forall \varepsilon>0\,\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. This again is a universal statement, so take an arbitrary $\varepsilon >0$).
$\color{blue}{\text{Let $\varepsilon >0$.}}$
(Now you wish to prove that $\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. 
(In words: there exists a positive real number $\delta$ with the property that for all real numbers $x$ such that $|x-c|<\delta$, $|(4-15x)-(4-15c)|<\varepsilon$).
This is an existential statement, one way to do it is to find $\delta >0$ such that $\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. This is what I suggest you do, find a $\delta$ that works). To do this, I suggest the thought of process below which I do in a scratch work box).

Scratch work: To prove that $\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$, we can begin by assuming we already have such $\delta$ (just to see what happens). 
So let $\delta>0$ be such that $\forall x\in \mathbb R(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon)$ and take an arbitrary $x\in \mathbb R$.
We have the conditional statement $|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon$.
Now it's useful to look at the consequent to see what kind of restrictions we need to impose on $\delta$ for this conditional statement to hold. It is almost always necessary to relate the consequent with the inequality in the antecedent ($|x-c|<\delta$).
The following holds: $|(4-15x)-(4-15c)|=15|x-c|$. One notices that in the consequent occurs the number $|x-c|$ which also occurs in the antecedent, thus relating the consequent to the antecedent. Recalling that $|x-c|<\delta$ one gets $|(4-15x)-(4-15c)|=15|x-c|<15\delta$
. This suggest that letting $\delta$ be such that $15\delta =\varepsilon$ (or equivalently $\delta=\dfrac \varepsilon {15}$) should work.
Now the scratch work is abandoned and we get back to the proof.

$\color{blue}{\text{Choose $\delta =\dfrac{\varepsilon}{15}$.}}$
(You wish to prove that $\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. It is a universal statement so you should take $x\in \mathbb R$ arbitrarily).
$\color{blue}{\text{Let $x\in \mathbb R$.}}$
(Now you want to prove the conditional statement $|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon$. In words: if $|x-c|<\delta$, then $|(4-15x)-(4-15c)|<\varepsilon$). The most forward way to do this is to assume the antecedent and try to prove the consequent).
$\color{blue}{\text{Suppose $|x-c|<\delta$.}}$
(Now you want to prove that $|(4-15x)-(4-15c)|<\varepsilon$).
$\color{blue}{\text{One has}}$
$$\color{blue}{\text{$\begin{align}|(4-15x)-(4-15c)|&=|4-15x-4+15c|\\ &=|-15x+15c|\\ &=15|-x+c|\\ &=15|x-c|<\varepsilon .\end{align}$}}$$
$\color{blue}{\text{Since $|x-c|<\delta$ and $\delta =\dfrac {\varepsilon}{15}$ it follows that}}$ $$\color{blue}{|(4-15x)-(4-15c)|=15|x-c|<15\delta=15\dfrac{\varepsilon}{15}=\varepsilon.\,\,\square}$$ 
This concludes the proof.

