My attempt:
Since $f$ has a limit at $x=a$. $\forall \epsilon>0$, $\exists \delta_0>0$ such that for all $x\in Dom(f(x))$ $$\delta_0>|x-a|\quad \text{implies}\quad \epsilon>|f(x)-L|$$
Why can't I just put $x=x-c$ because definiton says that it works for all x in domain, and the implication is already working.
You've written out, very nicely, the hypothesis that is assumed:
Hypothesis: For all $\epsilon > 0$ there exists $\delta_0 > 0$ such that if $x$ is in the domain of $f$ and if $\delta_0 > |x-a|$ then $\epsilon > |f(x)-L|$.
The key is to also write out the conclusion that you are supposed to prove, but to use a different symbol instead of $x$ to avoid a clash of dummy variables.
As written, the conclusion is $$\lim_{x \to (a+c)} f(x-c)=L $$ But let's change the dummy variable $x$ to $y$, so that the conclusion becomes $$\lim_{y \to (a+c)} f(y-c)=L $$ Now write this out in full using the definition of limits:
Conclusion: For all $\epsilon > 0$ there exists $\delta_0 > 0$ such that if $y-c$ is in the domain of $f$ and if $\delta_0 > |y-(a+c)|$ then $\epsilon > |f(y-c) - L|$.
Perhaps I'll stop here, you can probably see a simple variable substitution which makes the conclusion exactly equivalent to the hypothesis.
