Is my attempt at a proof valid? Warning: I have no formal education in these, so sorry if my attempt is horrible.
I've recently learnt the concept behind a limit, i.e. the epsilon-delta definition and it seems really cool. I've tried to 'prove' that the product of two continuous functions is also continuous, and would like help with fixing it up by pointing out its errors.
So, with the definition of continuity being that $\displaystyle\lim_{x\to\ a}f(x)=f(a)$ and $\displaystyle\lim_{x\to\ a}g(x)=g(a)$, I have to prove that $\displaystyle\lim_{x\to\ a}f(x)g(x)=f(a)g(a)$ for all real 'a'.Here's my attempt.
By the epsilon delta definition, for all $ε_f, ε_g >0$, there must exist $_f,_g$ such that:
For $x∈(a-_f,a+_f)$, $|f(x)-f(a)|<ε_f$
For $x∈(a-_g,a+_g)$, $|g(x)-g(a)|<ε_g$.
I have set $ε'=max(ε_f,ε_g)$ and $'=max(_f,_g)$ such that for x∈(a-',a+'):
$|f(x)-f(a)|<ε'$
$|g(x)-g(x)|<ε'$
Now, I must prove that for all $ε>0$, I can find  such that for $x∈(a-,a+)$.
$|f(x)g(x)-f(a)g(a)|<ε$
$\begin{align}|f(x)g(x)-f(a)g(a)|=&|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|\\≤&|f(x)||f(x)-f(a)|+|g(a)|g(x)-g(a)|\\<&|f(x)|ε'+|g(a)|ε'\end{align}$
I have set ε' to be determined later, and let the delta be . So if I show that I can find ε' for all ε>0 such that $|f(x)|ε'+|g(a)|ε'<ε$, I should be done with the proof since for $x∈(a-,a+), |f(x)-f(a)|<ε'$ and $|g(x)-g(a)|<ε'$.
$\begin{align}|f(x)|ε'+|g(a)|ε'=&|f(x)-f(a)+f(a)|ε'+|g(a)|ε'\\≤&[|f(x)-f(a)|+|f(a)|]ε'+|g(a)|ε'\\=&[ε'+|f(a)|]ε'+|g(a)|ε'\\=&(ε')^2+[|f(a)|+|g(a)|]ε'\end{align}$
I can set $(ε')^2+[|f(a)|+|g(a)|]ε'=ε$, which by the quadratic formula yields $ε'=\frac{-|f(a)|-|g(a)|+\sqrt{[|f(a)|+|g(a)|]^2+4ε}}{2}$, since the other solution is negative.
So, what I think I have proven is that for all ε>0, I can find ε' with the corresponding  being such that $x∈(a-,a+)$ will lead to $|f(x)g(x)-f(a)g(a)|<ε$
 A: Well There are few mistakes in your proof,
1.You need to prove $f(x)$ is bounded as a case and also take the case when either if $f(a)$,$g(a)$ are infinite.
2.Choose $\delta^{'}=\min({\delta_{f},\delta_{g}})$
Above everything it's a nice try!!!because you told you are a beginner so. Just fix these two points and also you need to be careful about doing the estimation of triangle inequality as mentioned in the comment below.
A: $\def\eps{\varepsilon}$You have had the idea, but writing a bit too much by trying to get "exact" $\eps$ and $\delta$ in various places. Also, the way you write the variables $\epsilon'$ and $\epsilon$ is confusing.

Here is a short proof that conceals most of the informal thoughts.
Proof. Let $\eps>0$. By continuity of $f$ and $g$ at $a$, let $\delta>0$ be such that for all $|x-a|<\delta$,
$$
\begin{align}
|f(x)-f(a)|<\min\left(\frac{\eps}{|f(a)|+|g(a)|+1},1\right),\\
|g(x)-g(a)|<\min\left(\frac{\eps}{|f(a)|+|g(a)|+1},1\right)\;.
\end{align}
$$
Then
$$
\begin{align}
|f(x)g(x)-f(a)g(a)|
&\le |f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|\\
&\le |f(x)||g(x)-g(a)|+|g(a)||f(x)-f(a)|\\
&\le \big(|f(x)-f(a)|+|f(a)|\big)|g(x)-g(a)|+|g(a)||f(x)-f(a)|\\
&\le \frac{|f(x)-f(a)|+|f(a)|+|g(a)|}{1+|f(a)|+|g(a)|}\eps\le\eps
\end{align}
$$
Q.E.D.

Here are the thoughts in terms of semi-formal languages on how one comes up with the proof above.
Let $\eps>0$. You want to find $\delta>0$ such that
$$
|f(x)g(x)-f(a)g(a)|<\eps
$$
whenever $|x-a|<\delta$. You have realized that the key step is to estimate the left-hand side as
$$
\begin{align}
|f(x)g(x)-f(a)g(a)|
&\le |f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|\\
&\le |f(x)||g(x)-g(a)|+|g(a)||f(x)-f(a)|\\
&\le \big(|f(x)-f(a)|+|f(a)|\big)|g(x)-g(a)|+|g(a)||f(x)-f(a)|
\end{align}\tag{1}
$$
so that you can exploit the continuity of $f$ and $g$ to control the right-hand side.
If you let $\delta>0$ be such that for all $|x-a|<\delta$,
$$
|f(x)-f(a)|<\eps,\quad |g(x)-g(a)|<\eps\;,\tag{2}
$$
then (1) implies that
$$
|f(x)g(x)-f(a)g(a)|\le (\eps+|f(a)|+|g(a)|)\eps\tag{3}
$$
Since $\eps$ is arbitrary, this already implies the limit you want. (This is how analysis is done in practice: one does not try to get a "perfect" $\eps$ on the right-hand side.)
But if you really want the right-hand side to be "pretty" as $\eps$, you can adjust (2). For instance, you can let $\delta>0$ be such that for all $|x-a|<\delta$,
$$
\begin{align}
|f(x)-f(a)|<\min\left(\frac{\eps}{|f(a)|+|g(a)|+1},\eps,1\right),\\
|g(x)-g(a)|<\min\left(\frac{\eps}{|f(a)|+|g(a)|+1},\eps,1\right)\;,
\end{align}\tag{2'}
$$
which implies by (1) that
$$
|f(x)g(x)-f(a)g(a)|
\le \frac{|f(x)-f(a)|+|f(a)|+|g(a)|}{1+|f(a)|+|g(a)|}\eps\le\eps\;.
$$
The idea here is that you can always adjust the right-hand side of (2') as small as you want by adding more (fixed) quantities in the "min" function. You do not need to solve a complicated quadratic equation to get the right $\delta$.
