Prove $f(x)$ is continuous for $\mathbb{R}$ 
Suppose that the domain of $f(x)$ is $\mathbb{R}$ and is continuous at $0$. Then if $f(x_1+x_2)=f(x_1)+f(x_2)$ for all real values $x_1$ and $x_2$, show that $f(x)$ is continuous for all reals.

We know that, $$\lim_{x\to0}f(x)=f(0)$$
From that I wanted to do something like, 
Let $x_1+x_2=0$ and show that $$\lim_{x\to0}f(x_1+x_2)=f(0)$$
Am I even going about this the right way? To show that $f(x)$ is continuous on $\mathbb{R}$, I also thought to maybe relate it to the Intermediate Value Theorem. I am stuck, and I also can not find a forsure answer on whether given a domain of all real values, does it mean that $f$ is continuous on $\mathbb{R}$? I am Lost! I can not use derivatives, or integration as I have not learned either. Only using the basic limit theorems of continuity or the the intermediate value theorem.
UPDATE: I found an $\epsilon-\delta$ proof! I posted it below as the answer!
 A: Consider the sequence $x_n \to x $ for any $x \in \mathbb{R}$. Then the sequence $(x_n-x) \to 0$. Now we use linearity:
$$
|f(x_n)-f(x)|=|f((x_n-x)-0)|=|f(x_n-x)-f(0)| \to 0
$$
The intermediate value theorem only works if already know that your function is continuous. Consult your textbook/class if you have a basic lack of understanding, they can handle and explain your problems better and thats what they are there for!
A: My making $x_1=x_2 =0$ we get $f(0)=0$. Then $0=f(0)=f(x + (-x))=f(x) + f(-x)$, therefore $f(-x)=-f(x), \forall x$.
Now let $x$ arbitrary and $x_n \rightarrow x$. Then $f(x_n) - f(x) = f(x_n) + f(-x)=f(x_n -x)$. Because $x_n - x \rightarrow 0$ and $f$ continous at $0$ it follows $lim_{n \rightarrow \infty} f(x_n -x) = f(0) = 0$ therefore $lim_{n \rightarrow \infty} f(x_n) - f(x) = 0$
A: A one line answer:
$$\lim_{h \to 0}f(x + h) = \lim_{h \to 0}f(x) + f(h) = f(x) + f(0) = f(x + 0) = f(x)$$

I looked at other answers and I could not understand the need for all the extra stuff done there. The problem is a relatively simple one which requires you to use just the definition of continuity and nothing more. There is no need to calculate $f(0)$ in particular.
A: $f(x)=f(x+0)=f(x)+f(0) \implies f(0)=0$ and
$f(0)=f(x-x)=f(x)+f(-x)=0 \implies f(-x)=-f(x)$ $\forall x$
$\epsilon-\delta$ Proof:
Let $\epsilon>0$ and choose $\delta>0$ such that,
$$|t-0|<\delta \implies |f(t)-f(0)|<\epsilon$$
i.e
$$|t|<\delta \implies |f(t)|<\epsilon$$
Let $a$ be any real number.
Assume $t=x-a$. Then,
$$|x-a|<\delta \implies |f(x-a)|=|f(x)-f(a)|<\epsilon$$
Therefore, $\lim_{x\to a}f(x)=f(a)$ $\forall a\in \mathbb{R}$. i.e. $f$ is continuous on$(-\infty,\infty)!$
A: First we have, for all $x$, $f(x)=f(0+x)=f(0)+f(x)$, which shows that $f(0)=0$.
Now, for all $x$, $f(0)=f(x-x)=f(x)+f(-x)$, so $f(-x)=-f(x)$.
Finally, for any $x_0$ in $\mathbb{R}$, we have $\lim\limits_{x\to x_0} f(x) = \lim\limits_{x\to 0} f(x_0-x) = f(x_0)+\lim\limits_{x\to 0}-f(x)=f(x_0)$.
