You have made one common arithmetic mistake and two common conceptual mistakes.
The arithmetic and first conceptual mistake are mistakes of algebra. The second conceptual mistake is the first mistake that's actually related to calculus.
The arithmetic mistake is the usual one regarding square roots and forgetting that negative numbers exist — the detail here is a bit complicated; the easiest method would be to go back to your precalculus text and review solving inequalities.
Henceforth, I will assume all the variables are such that your work is correct.
The first conceptual mistake is the usual mistake of proving the converse of what you are trying to solve. You have proven
If $\lvert x^2 - c^2\rvert < \epsilon$, then $\lvert x-c\rvert < \delta$
but what you are trying to prove is
If $\lvert x - c\rvert < \delta$, then $\lvert x^2 - c^2\rvert < \epsilon$
This is a common mistake, because you're taught to solve problems backwards; working backwards is an effective means of simplifying problems, but once you have what you believe to be a solution, you have to turn around and prove that it is so. Again, you can see examples of this in your precalculus class, where you solve equations like $\sqrt{x+2} = x$ but get "spurious" solutions; in this case, $x=-1$.
The other conceptual mistake is that you are trying hard to solve the problem exactly. One of the great things that makes calculus "simple" is that we don't have to solve intervals like this exactly — we can usually make all sorts of simplifications to make our life easier.
The usual trick here is to observe that
$$ \lvert x^2 - c^2\rvert = \lvert x-c\rvert \lvert x+c\rvert $$
so if $\lvert x-c\rvert < \delta$, then
$$ \lvert x^2 - c^2 \rvert = \lvert x-c\rvert \lvert x+c\rvert < \delta \lvert x+c\rvert $$
To deal with the $\lvert x+c\rvert$ term, we simply observe that $x$ is near $c$, so $\lvert x+c\rvert$ shouldn't be much bigger than $\lvert 2c\rvert$.
So, we add a second goal: in addition to wanting to prove
$$ \lvert x - c\rvert < \delta \implies \lvert x^2 - c^2\rvert < \epsilon $$
we also want to have
$$ \lvert x - c\rvert < \delta \implies \lvert x + c\rvert < \lvert 2c\rvert + 1 $$
The latter isn't hard to get: it's enough to pick $\delta \leq 1$.
Thus, if we have both $\lvert x - c\rvert < \delta$ and $\delta < 1$, then we have
$$ \lvert x^2 - c^2\rvert = \lvert x-c\rvert \lvert x+c\rvert < \delta (\lvert 2c\rvert + 1) $$
So if we can additionally choose $\delta$ so that $\delta (\lvert 2c\rvert + 1) \leq \epsilon$, then we have what we need!
Choosing $\delta = \min\left( \frac{\epsilon}{\lvert 2c\rvert + 1}, 1 \right)$ works to satisfy both of the conditions we needed for $\delta$.
This proof is not correct. The key error is when you say
Since this must be true for any $\epsilon>0$, it must be true for some $\epsilon>|f(cx)-f(x)|+|f(x)-L|$.
You're trying to prove that for every $\epsilon>0$, there exists $\delta>0$ such that $0<|x|<\delta \implies |f(cx)-L|<\epsilon$. So you're not allowed to choose whatever $\epsilon$ you want; $\epsilon$ is given to you ahead of time. It is true that for any particular $x$, there exists some $\epsilon$ such that $\epsilon> |f(cx)-f(x)|+|f(x)-L|$. But this isn't any use, since you don't get to choose $\epsilon$. (Moreover, you would need this inequality to hold simultaneously for every $x$ such that $0<|x|<\delta$, and it is not clear how you are getting that.)
Best Answer
You have all the right ideas, and your proof could be considered correct, given some benefit of the doubt. But in early analysis courses, it's all about being completely precise, therefore I have here written a more streamlined proof, which explain every step meticulously.
We want to prove that $\lim_{x\rightarrow a} f(x)g(x) = \infty$. In other words, given a positive $M$, we want to find $\delta >0$ such that $f(x)g(x) > M$ for all $x$ with $|x-a|<\delta$. Now that we have written down exactly what to prove, here is how I would prove it.
Let $M>0$. Since we want to show that the limit is positive infinity, let us show that $g(x)$ stays positive if x is close enough to $a$. Since $c>0$, we have $\varepsilon = \frac{c}{2}>0$. Therefore, we can choose a $\delta_1>0$ such that $f(x)>\frac{c}{2}>0$, given that $|x-a|<\delta_1$
Now let $N = \frac{M}{\frac{c}{2}}>0$. Since $\lim_{x \rightarrow a}f(x) = \infty$, we can choose $\delta_2>0$ such that $f(x)>N$ when $|x-a|<\delta_2$. Now it follows that if we set $\delta = \min(\delta_1, \delta_2)$, we get for $|x-a|<\delta$ that
$f(x)g(x)>N\frac{c}{2}$=M
as desired.