I want to know every single bit there is to understand in this following proof I'm ashamed of myself for not answering my professor when he asked if I found the proof myself. I gave him the impression that I did which was still not true. So I want to make it right. 
It's about the proof of the product of the functions $f$ and $g$ is continuous. I found the proof here, it's on the 1st page on the bottom of the page. 
I'd like to know why he/she starts with setting prerequisites like $\epsilon \lt 1$ and $M = \max(f(a), |g(a)|,1)$. I certainly don't understand why to put $M$ like that. 
And the second sentence I understand because first of all it's defining what we already know about continuity and then setting an assumption,ok. 
But then the inequality $f(x) \lt f(a) + \dfrac{\epsilon}{3M} \le M + \dfrac{\epsilon}{3M}$ Why exactly did he/she do that? 
Thanks for making things right. 
 A: First of all, note that $a$ is chosen beforehand, so you are allowed to use anything directly dependent on $a$. Now the inequality:
$$|(fg)(x) − (fg)(a)| \leq |f(x)||g(x) − g(a)| + |g(a)||f(x) − f(a)|$$
we are able to estimate $|f(x) − f(a)|$ and $|g(x) − g(a)|$ by $\varepsilon$, but to do this correctly, we must also absolve $|f(x)|$ and $|g(a)|$. Note that the latter one is a fixed number, since $a$ was chosen first. We need to estimate $|f(x)|$. This can be done by using the reverse triangle inequality: $||f(x)| − |f(a)||\leq |f(x) − f(a)|$, which implies that $|f(x)| − |f(a)|\leq |f(x) − f(a)|$ and thus $|f(x)|\leq|f(a)|+|f(x) − f(a)|$. We see that both can be estimated.
Now for the estimation:
we want some number greater than both $|f(a)|$ and $|g(a)|$ (so the multiplicative inverse is small), but also greater than $1$ to prevent dividing by $0$, so we can choose some $M=\max(|f(a)|, |g(a)|,1)$:
$$\color{red}{|f(x)|}|g(x) − g(a)| + \color{green}{|g(a)|}|f(x) − f(a)|< \color{red}{|f(x)|}\varepsilon + \color{green}{|g(a)|}\varepsilon\leq\color{red}{(M+\varepsilon)}\varepsilon+\color{green}{M}\varepsilon$$
Now of course we want this all estimated by only one $\epsilon$, so in the beginning he chooses $\varepsilon=\frac{\epsilon}{3M}$, where we include the $M$ to absolve the $M$'s and $3$ to cleanly estimate the three $\varepsilon$-terms. We see that
$$(M+\varepsilon)\varepsilon+M\varepsilon=\frac{\epsilon}{3}+\frac{\epsilon^2}{9M^2}+\frac{\epsilon}{3}$$
Now without loss of generality (because these proofs are always about small $\epsilon$'s) we let $\epsilon<1$ to cleanly estimate the middle term, so $$\frac{\epsilon^2}{9M^2}=\epsilon\frac{\epsilon}{9M^2}<\frac{\epsilon}{9M^2}\leq\frac{\epsilon}{9}$$
where we remember that $M\geq 1$, so $\frac{1}{M}\leq 1$.
And so we end up with
$$\frac{\epsilon}{3}+\frac{\epsilon^2}{9M^2}+\frac{\epsilon}{3}< \epsilon$$
I find his proof rather difficult. It can be done much easier.
Also, this looks like a lot, but later on you will get used to this kind of thinking (and everything turns trivial).
A: I'm typing this for fun.  The question asks for explanation of the linked proof.  I'll first explain the proof, before using it to analyse that proof, so that OP knows the role of $M$, "$\epsilon<1$" and $\epsilon/3$ in that proof.
Criticism of the linked proof
To understand elementary epsilon-delta proofs in calculus, I don't follow the literal order--they are put in such way so as to get a logically correct answer.  Logically correct things may not be instructive to mathematicians, such as the proof of $1+1=2$ and the proof of Zorn's Lemma.
The use of $\epsilon/3$ in the linked proof is not suggested for students, especially in tests or exams.  You can easily mess up things under stress and pressure---It is logically fine to put $M\epsilon$, where $M$ is a constant.  This practice is common in more advanced math.  In fact, the author tries to change $M\epsilon$ to $\epsilon$.  This adds no rigors to the proof, but renders the arguments too complicated for revision and reading---this motivates me to give such criticism to that proof.
Step-by-step explanation of the proof
Thinking from the reverse enables you to assimilate the proof, so that you can respond to similar questions in the future -- that's the fool-proof way.


*

*Fix $\epsilon > 0$

*Jump to $|(fg)(x)-(fg)(a)|<\epsilon$ skip the settings on $\delta$.  This is our goal.

*Think about $f(x)\underset{x\to a}\longrightarrow f(a)$ and $g(x)\underset{x\to a}\longrightarrow g(a)$ in (vague) words: "when $x$ is near $a$, ...".  Use this to get $\delta' := \delta_f \wedge \delta_g$ so that you may use $|f(x)-f(a)|<\epsilon$ and $|g(x)-g(a)|<\epsilon$ at the same time.  (Write $\delta_1,\delta_2$ if you want.)

*To use the inequalities in step (3) to finish our goal in (2), we need to insert either $f(x)g(a)$ or $f(a)g(x)$.  WLOG, we stick with the author's choice: $f(x)g(a)$.

*To balance the effect of (4), we put another $f(x)g(a)$ into LHS of (2) but with opposite sign, so that it becomes $|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|$

*From this, it's obvious that we have to apply the triangle inequality and factor out some terms to get $|f(x)||g(x) − g(a)| + |g(a)||f(x) − f(a)|$.

*Apply (3): intuitively when $x$ is "near" $a$, $|f(x)|$ is "near" $|f(a)|$.  Draw a (vertical) straight line to see the geometric meaning of $|f(x)-f(a)|<\epsilon$. (Distance between $f(x)$ and $f(a)$ is less than $\epsilon$.)  From this, we easily get $|f(x)|<|f(a)|+\epsilon$.  Then (6) is bounded above by \begin{align}
& |f(x)||g(x)-g(a)|+|g(a)||f(x)-f(a)| \\
&\le (|f(a)|+|f(x)-f(a)|)|g(x)-g(a)|+|g(a)||f(x)-f(a)| \\
&= \color{blue}{|g(a)||f(x)-f(a)|} + \color{red}{|f(a)||g(x)-g(a)|} + |f(x)-f(a)| |g(x)-g(a)| \\
&< (|f(a)|+\epsilon)\epsilon + |g(a)|\epsilon \\
&= (|f(a)|+|g(a)|) \epsilon + \epsilon^2.
\end{align}

*Intuitively $\epsilon$ is "small", (so $\epsilon^2$ is even "smaller"), so we're "done".  


*

*Logically in (1), we have for all $\epsilon > 0$ in the very definition of limit, so we can't discard the square $^2$ by saying that $\epsilon^2 \le \epsilon$.  To fix this logical problem, we change (1) to $0 < \epsilon \le 1$.

*Since $\epsilon$ is arbitrary above, this doesn't affect the validity other established inequalities.  So the existence of $\delta$ responding to the question has been proven for the case $0 < \epsilon \le 1$.

*If $\epsilon > 1$, apply the results proven for $\epsilon' = 1$: $$\forall\,\epsilon>1, \exists\,\delta > 0 \dots |\cdots| < 1 < \epsilon$$



Explanation of that proof using the proof
As another answer points out, his professor want the RHS
$$ \color{blue}{\epsilon |g(a)|} + \color{red}{\epsilon |f(a)|}
+ \epsilon^2 $$
"all estimated by only one" $\epsilon$.  To do so, he set another "smaller" $\epsilon$ in step (3) (i.e $\epsilon$ divided a "large" number $M$), so that
$$\color{blue}{|f(x)-f(a)||g(a)|<\frac{\epsilon}{3} \iff |f(x)-f(a)| < \frac{\epsilon}{3|g(a)|}}$$
and
$$\color{red}{|g(x)-g(a)||f(a)| < \frac{\epsilon}{3} \iff |g(x)-g(a)| < \frac{\epsilon}{3|f(a)|}}$$
if $f(a)\ne0$ and $g(a)\ne0$.
We set $M \ge 1$ to fix this problem, so that $M = \max(|f(a)|,|g(a),1)$.  It's easy to check that $|f(x)-f(a)| |g(x)-g(a)| < \dfrac\epsilon3$.
Final Remark: Comparison of my answer than the alternate answer


*

*The alternative answer uses the fact that $3\times3=9$, as in that proof.  Mine has marginally simpler arithmetic calulations: I only use $1+1=2$ and $1\times3=3$.

*My answer has more bullet points and numbered list.

