Construction of function from given condition on function

Question

Let $f$ be defined on $\mathbb R$ and Assume that there exist at least one point $x_0$ in $\mathbb R$ at which $f$ is continuous .Suppose also that for every $x$ and $y$ in $\mathbb R$ ,$f$ satisfies following equation
$f(x+y)=f(x)+f(y)$
Prove that $\exists a$ which is constant such that $f(x)=ax$ for all $x$

My attempt $f(0)=0$ by substitution .As $f$ is continous at $x_0$ then there exist sequnce $x_n$ such that $x_n \to x_0$ and $f(x_0)=\lim_{n\to \infty}f(x_n)$
Also I can say that $\forall \epsilon >0 \exists \delta>0$ such that there is $h \in \mathbb R$ such that $|f(x_0+h)-f(x_0)|<\epsilon$ whenever $|h|<\delta$
$|f(x_0+h)-f(x_0)|=|f(h)|<\epsilon$ whenever $|h|<\delta$
As $f(0)=0$ so $f$ is continuous at $0$.
For integer $m$ ,$f(m.1)=m f(1)$.
For rational point $\frac mn$.
$f(\frac mn)=mf(\frac 1n)$
$f(1)=f(m(\frac 1m))=mf(\frac 1m)$.SO $\frac {f(1)}{m}=f(\frac 1m)$
Therefore $f(\frac mn)=\frac {m}{n}f(1) $
Consider sequnce $x_n$ $\in \mathbb Q$ such that converging to $x \in \mathbb R$ so $f(x)=\lim_{n\to \infty}f(x_n)$=$\lim_{n\to \infty}x_n f(1)$=xf(1) Done Thanks for Hint.

Answer 1

f(0)=0 .As f is continous at $x_0$ then there exist sequnce $x_n$ such that $x_n \to x_0$ and f($x_0$)=$\lim_{n\to \infty}f(x_n)$
Also I can say that $\forall \epsilon >0 \exists \delta>0$ such that there is h $\in R$ such that |f($x_0+h$)-f($x_0$)|<$\epsilon$ whenever |h|<$\delta$
|f($x_0+h$)-f($x_0$)|=|f($h$)|<$\epsilon$ whenever |h|<$\delta$
As f(0)=0 so f is continuous at 0.
For integer m ,f(m.1)=m f(1).
For rational point m/n.
f(m/n)=mf(1/n)
f(1)=f(m(1/m))=mf(1/m).SO f(1)/m=f(1/m)
Therefore f(m/n)=$\frac {m}{n}$f(1)
Consider sequnce $x_n$ $\in Q$ such that converging to x $\in R$ so f(x)=$\lim_{n\to \infty}f(x_n)$=$\lim_{n\to \infty}x_n f(1)$=xf(1)