Here is Prob. 12, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $f \colon \mathbb{R} \to \mathbb{R}$ be such that $f(x+y) = f(x) + f(y)$ for all $x, y$ in $\mathbb{R}$. Assume that $\lim_{x \to 0} f = L$ exists. Prove that $L = 0$, and then prove that $f$ has a limit at every point $c \in \mathbb{R}$. [Hint: First note that $f(2x) = f(x) + f(x) = 2 f(x)$ for $x \in \mathbb{R}$. Also note that $f(x) = f(x-c) + f(c)$ for $x, c$ in $\mathbb{R}$.]
My Attempt:
As $\lim_{x \to 0} f (x) = L$, so, given any real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$ \lvert f(x) - L \rvert < \varepsilon \tag{1} $$ for all $x \in \mathbb{R}$ for which $$ 0 < \lvert x - 0 \rvert < \delta; \tag{2} $$ therefore, we can also conclude that, for all $x \in \mathbb{R}$ for which $$ 0 < \lvert x - 0 \rvert < \frac{\delta}{2} $$ holds, for those $x$ we also have $$ 0 < \lvert 2x - 0 \rvert < \delta, $$ and so $$ \lvert f(2x) - L \rvert < \varepsilon, $$ and thus it follows that $$ \lim_{x \to 0} f(2x) = L $$ also. But as $f(2x) = 2 f(x)$ for all $x \in \mathbb{R}$, so we must have $$ \lim_{x \to 0} f(2x) = 2 \lim_{x \to 0} f(x), $$ that is we must have $$ L = 2L,$$ which in turn implies that $L = 0$. Hence $$ \lim_{x \to 0} f(x) = 0 $$ holds.
Am I right? Is this part of the proof correct and clear enough?
Now for any real numbers $c, x$, we find that $$ f(x) = f(x-c + c) = f(x-c) + f(c), $$ and so $$ f(x) - f(c) = f(x-c). $$ Now if $x= c$, then we obtain $$ f(c) - f(c) = f(c-c) $$ or $$ f(0) = 0. $$ Then using (1) and (2) above, we find that, for every real number $\varepsilon > 0$, there exists a real number $\delta > 0$ such that $$ \lvert f(x) - f(c) \rvert = \lvert f(x-c) \rvert < \varepsilon $$ for all $x \in \mathbb{R}$ for which $$ 0 < \lvert x-c \rvert < \delta. $$ Therefore we can conclude that $$ \lim_{x \to c} f(x) = f(c). $$
Is this proof sound enough in each and every detail hereof?
Proof
Since $$f(x)+f(y)=f(x+y),~~~\forall x,y \in \mathbb{R},\tag1$$ hence $$f(x)+f(x)=f(x+x),~~~\forall x\in \mathbb{R},\tag2$$ namely $$2f(x)=f(2x),~~~\forall x\in \mathbb{R}.\tag3$$ Let $x \to 0$ and take the limits of the both sides of $(3)$. We obtain $$2L=2\lim_{x \to 0}f(x)=\lim_{x \to 0}2f(x)=\lim_{x \to 0}f(2x)=L,$$which shows that $$L=0.$$
Moreover, from $(1)$ we may obtain $$f(c)+f(x-c)=f(x),~~~\forall x,c \in \mathbb{R},\tag4$$where $c$ is any a constant. Likewise, let $x \to c$ and take the limits of both sides of $(4).$ We obtain $$f(c)+0=\lim_{x \to c}f(c)+\lim_{x \to c}f(x-c)=\lim_{x \to c}[f(c)+f(x-c)]=\lim_{x \to c}f(x),$$namely $$\lim_{x \to c}f(x)=f(c),$$which shows that $f(x)$ has a limit $f(c)$ at $x=c$, in another word, it is continuous at $x=c.$ We are done!