I'm working on Spivak and I've come across this problem.
Suppose that $f$ satisfies $f(x+y) = f(x) + f(y)$, and $f$ is continuous at $0$. Prove that $f$ is continuous at a for all a.
OK, then the working definition of continuity: given $\epsilon>0$ there exists a $\delta>0$ such that if $x$ is in $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$. In other words, $\lim_{x \to a} f(x) = f(a)$.
So I've gotten to this point:
Since $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{R}$ then $f(x+0) = f(x) + f(0)$ so $f(0) = 0$.
Now, $\lim_{h \to 0} f(x+h) - f(x)$
= $\lim_{h \to 0} f(x) + f(h) - f(x)$
= $\lim_{h \to 0} f(h)$
= $\lim_{h \to 0} f(h) - f(0) = 0.$
I don't know how to conclude now that $\lim_{x \to a} f(x) = f(a)$ for all a.
edit: A hint is all that's necessary, I'd still like to try to work through the exercise myself if I can.
You proved that, for each $x\in\mathbb R$,$$\lim_{h\to0}f(x+h)-f(x)=0,$$which is the same thing as asserting that$$\lim_{h\to0}f(x+h)=f(x).$$And you're done! This proves that $f$ is continuous at $x$.