$f: \mathbb R \rightarrow \mathbb R, f(x+y)=f(x)+f(y), \forall x,y \in \mathbb R$
I can see that $f(2x)=f(x)+f(x)=2f(x)$ and $f(x-c)=f(x)-f(c)$, also that $f(0)=0$.
Nothing is mentioned about continuity of the function, so I cannot use the assumption of continuity.
To prove that $L=0$, I need to show that $|f(x)|<\varepsilon$ when $|x-c|<\delta$
I don't know how to go about it. Please help!
By induction show that $f(nx)=nf(x)$, hence $f(\tfrac1n)=\frac1nf(1)\to 0$.
Hint: if $f$ is identically $0$, you are done. Suppose it is not, i.e. there exists $x^\ast$ such that $f(x^\ast)\neq 0$.
You can show that $f(x^\ast) = 2^n f\!\left(\frac{x^\ast}{2^n}\right) $ for all $n \geq 0$ (by induction). And the assumption $\lim_0 f = L$ implies that $f\!\left(\frac{x^\ast}{2^n}\right)\xrightarrow[n\to\infty]{} L$ (this is not assuming continuity, only the limit: nothing says $f(0)=L$). Can you conclude?
This is the standard approach to such problems: Assume $L\neq0$. Show, by using the additive property, that given a small enough $\epsilon$ (like $\frac{|L|}{10}$, for instance) there is no $\delta$ that works.
It is easy to show that $f(x) = f(1)x$, whenever $x \in \mathbb{Q}$.
Now since you are given that $\lim_{x \to 0}f(x) = L$ exists, we can consider convergence to $0$ along the sequence $x_n=1/n$, i.e., we have $\lim_{n \to \infty} f(1/n) = L$. This implies that $L = \lim_{n \to \infty} \frac{f(1)}n = 0$.
