Questions about continuity in the function $f(x+y)= f(x)+f(y)$ for each $x,y \in \Bbb{R}$. $f: \Bbb{R}\rightarrow\Bbb{R}$ is continuous in zero, s.t $f(x+y)= f(x)+f(y)$ for each $x,y \in \Bbb{R}$.
A. calculate $f(0)$ and prove that $f(-x)=-f(x)$
B. prove that $f$ is continuous in $\Bbb{R}$ 
for A, I reached that the limit when $x$ goes to zero must be equal to the value $f(0)$, by the definition of continuity, which means that $x+y$ must go to zero, and that way x can go to $-y$ which is the same thing, then I put $-y$ in the function and see what value I get, at the end i got: 
$\lim \limits_{x \to -y}$$f(x+y) = f(-y)+f(y)$.
and I know from the information I have about the function $f(x+y)= f(x)+f(y)$ that $f(y)+f(-y) = f(0)$, but didn't know how to continue from here. and I don't have any direction for B. any kind of help would be appreciated. 
 A: $f(0)=0$ is shown as indicated by @user109899.
For B):
$$\lim_{y\to 0}f(x+y)=\lim_{y\to 0}f(x)+\lim_{y\to 0}f(y)=f(x)+f(0)=f(x) $$
by the hypothesis of continuity at $0$ and point A.
A: If $y=0$ we get
$$f(x)=f(x)+f(0)\quad\Longrightarrow\quad f(0)=0$$
If $y=-x$ we get
$$0=f(0)=f(x)+f(-x)\quad\Longrightarrow\quad -f(x)=f(-x)$$
Now, since $f$ is continuous at $x=0$, we know that
$$\forall \epsilon>0\quad\exists\delta>0\quad \text{such that}\quad |x|<\delta \quad\Longrightarrow\quad|f(x)|<\epsilon$$
In particular, fix any $x\in\mathbb{R}$, now for all $y\in \mathbb R$ we have
$$|y-x|<\delta \quad\Longrightarrow\quad|f(y)-f(x)|=|f(y-x)|<\epsilon$$
Thus, $f$ is continuous at $x$.
A: If $f$ is locally bounded and $f(x+y)=f(x)+f(y)$, then $f(n x)=n\,f(x)$ for any $n\in\mathbb{N}$, so
$$|f(x/n)|=\frac{1}{n}|f(x)|$$
gives that $f(0)=0$ and $f$ is continuous in zero. From $0=f(0)=f(x)+f(-x)$ we get that $f$ is an odd function, and from $|f(x+y)-f(x)|=|f(y)|$ we get that $f$ is everywhere continuous, by considering $y\to 0$. Along the same lines, we can prove that if $f$ is locally bounded and $f(x+y)=f(x)+f(y)$, then $f(x)=\alpha x$ for some $\alpha\in\mathbb{R}$.
