Show a function for which $f(x + y) = f(x) + f(y) $ is continuous at zero if and only if it is continuous on $\mathbb R$

First observe that $f(0) = f(0) + f(0)$, so that $f(0) = 0$.

Now suppose $f$ is continuous at $0$. Let $x \in \mathbb{R}, \epsilon > 0$. Let $\delta > 0$ be such that $|f(t)| < \epsilon$ whenever $|t| < \delta$.

If $|y - x| < \delta$, then setting $t = y -x$ we have $|f(y) - f(x)| = |f(t)| < \epsilon$, thus completing the proof.

Extra comments: Of course we can prove much more than continuity if $f$ is continuous at zero: it must be of the form $f(x)= \alpha x$, for some $\alpha \in \mathbb{R}$.

There's a related fact (harder to prove, but not that hard): Suppose $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x) + f(y)$, and suppose $f$ is not continuous. Then the graph of $f$, $\Gamma(f) = \{(x,f(x)) | x \in \mathbb{R}\}$, is dense in $\mathbb{R}^2$.

Hints:

If $f(x)$ is continuous at 0, then we have $\lim_{x\to 0}f(x)=0$.

Then $\lim_{x\to 0}f(x_0+x)=\lim_{x\to 0}(f(x_0)+f(x))=f(x_0)+\lim_{x\to 0}f(x)=f(x_0), $which shows that $f(x)$ is continuous at $x_0$.

First let $x=y=0$, we have $f(0)=2f(0)$ which means $f(0)=0$.
For any $a\in R$ and a given $\epsilon>0$, because f is continuous at 0, there exist $\delta>0$ s.t. $|x|<\delta$ implies $|f(x)-f(0)|=|f(x)|<\epsilon$. Now consider the same $\delta$, if $|x-a|<\delta$, we have $$|f(x)-f(a)|=|f(x-a)|<\epsilon$$, therefore f is continuous at $a$.