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
Fix $\varepsilon > 0$ and $a$.
From the definition of differentiation we have $$ \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \varepsilon $$
for an appropriately chosen $\delta > 0$.
Multiply both sides by $|x - a|$ to get: $$ \left|f(x) - f(a) - (x - a)f'(a)\right| < |x - a| \varepsilon $$
Using $\left||x|-|y|\right| \le |x - y|$ we have:
$$ \left|f(x) - f(a)\right| - |x - a| \cdot \left|f'(a)\right| < |x - a| \varepsilon $$
Rearrange to get: $$ \left|f(x) - f(a)\right| < (\left|f'(a)\right| + \varepsilon) \cdot |x - a| $$
Since $f'(a)$ and $\varepsilon$ are both fixed, you can make $|f(x) - f(a)|$ as small as you want by making $|x - a|$ smaller and smaller. Thus, the function is continuous at $a$.
To prove this formally, pick any $\hat{\varepsilon}$ (different from $\varepsilon$ fixed at the beginning and used with the differentiation definition). Pick $\hat{\delta} = \min\left(\delta, \frac{\hat{\varepsilon}}{\left|f'(a)\right| + \varepsilon}\right)$. Clearly:
$$ |x - a| < \hat{\delta} \Rightarrow \left|f(x) - f(a)\right| < \hat{\varepsilon} $$