Let $f$ be a function defined on all of $\mathbb{R}$ that satisfies the additive condition $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. I have already proved that $f(0)=0$ and $f(-x)=-f(x)$ for all $x \in \mathbb{R}$, therefore I will take these facts as given.
I want to show that if $f$ is continuous at $x=0$, then $f$ is continuous at every point in $\mathbb{R}$. I will use the following definition of continuity: A function $f: A \rightarrow \mathbb{R}$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-c|<\delta$ it follows that $|f(x)-f(c)| < \epsilon$.
My Proof Attempt: Let $\epsilon>0$. Since $f$ is continuous at $x=0$, there exists a $\delta > 0$ such that $|x – 0| = |x| < \delta$ implies $|f(x) – f(0)| < \epsilon$. Since $f(0) = 0$, we have $$|x| < \delta \Rightarrow |f(x)| < \epsilon. \hspace{1em} (1)$$
Let $y_0 \in \mathbb{R}$ with $y_0 \neq 0$. We want to show that $f$ is continuous at $y_0$. We have (for $a \in \mathbb{R}$) $$|a – y_0| < \delta \Rightarrow |f(a – y_0)| < \epsilon \hspace{1em}(2)$$
by (1). The additive condition gives $|f(a-y_0)| = |f(a) – f(y_0)|$. Substituting this into (2) gives $$|a – y_0| < \delta \Rightarrow |f(a) – f(y_0)| < \epsilon$$
So, by definition of continuity, $f$ is continuous at $y_0$.
Best Answer
Your proof looks fine to me, but could be a little better structured. Here's how I would do it:
Proof: As you have already shown, $f(0)=0$ and $f(-x)=-f(x)$ for all $x\in\mathbb{R}$.
Let $x_0\in\mathbb{R}$ and $\epsilon>0$. Since $f$ is continuous at $x=0$ and $f(0)=0$, there exists $\delta>0$ such that for each $x\in\mathbb{R}$, we have that $$|x|<\delta\implies |f(x)|<\epsilon.$$ Now if $x\in\mathbb{R}$ with $|x-x_0|<\delta$, we have that $$|f(x-x_0)|=|f(x)-f(x_0)|<\epsilon$$ using your above observations. Since $\epsilon>0$ was chosen arbitrarily, this shows directly from the definition that $f$ is continuous at $x_0$ (for any $x_0\in\mathbb{R}$), and we are done.