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

Suppose that $f: \mathbb R \to\mathbb R$ satisfies $f(x + y) = f(x) + f(y)$ for each real $x,y$.

Prove $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb R$.

Proof: suppose $f$ is continuous at zero. Then let $R$ be an open interval containing zero. Then $f$ is continuous at zero if and only if $f(x) \to f(0)$ as $x \to 0$. Then $|f(x) - f(0)| < \epsilon$.

Can anyone please help me? I don't really know how to continue. Thank you.

Answer

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 $a$.