I'm assuming your true concern is understanding $\epsilon$,$\delta$-proofs. We first go over the definition of continuity at a point, then give an example.
The Definition
Let $f$ be a function from some subset of $\mathbb{R}$, call it $D$ for domain, to another subset of $\mathbb{R}$, call it $C$ for codomain. That is,$$f:D\subseteq\mathbb{R}\to C\subseteq\mathbb{R}$$
Let $a$ be an arbitrary element of the domain. That is, $$a\in D$$
$f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in D$, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
An Example
Let $f:[3,\infty)\to \mathbb{R}$ be defined by
$$f(x)=\sqrt{2x-6}$$
for every $x\in[3,\infty)$. (It is important to define the function explicitly; that is, the domain is relevant for this proof.)
$f$ is continuous at $4$.
Proof. To prove $f$ is continuous at $4$, we will show that for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$.
Let $\epsilon>0$ be arbitrary. Notice for each $x\in[3,\infty)$,
$$|f(x)-f(4)|=\sqrt{2}|\sqrt{x-3}-1|=\sqrt{2}\left|\frac{x-4}{\left(\sqrt{x-3}\right)+1}\right|<\sqrt{2}|x-4|\tag{1}$$
Define $\delta\overset{\text{def}}{=}\dfrac{\epsilon}{\sqrt{2}}$.
It is important to understand why we defined $\delta$ this way (in relation to $\epsilon$).
Anyway, we continue with the proof.
For each $x\in[3,\infty)$, if $|x-4|<\delta$, then
$$|f(x)-f(4)|\overset{(1)}{<}\sqrt{2}|x-4|<\sqrt{2}\cdot\delta=\epsilon$$
Recall $\epsilon>0$ was arbitrary. Therefore for each $\epsilon>0$, there exists $\delta>0$ (namely $\epsilon/\sqrt{2}$), such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$. Therefore $f$ is continuous at $4$.$\square$
Best Answer
This definition would not be very useful. For example, any bounded function would be continuous. Take $f\colon\Bbb R\to\Bbb R$ where $f(x)=1$ if $x\in\Bbb Q$ and $f(x)=0$ if $x\notin\Bbb Q$. For any $c\in\Bbb R$ and any $\delta>0$, we have $|x-c|<\delta\implies |f(x)-f(c)|\le 1$. So $f$ is continuous at every point.