Is every monotonic additive function $f \colon \mathbb{R} \to \mathbb{R}$ continuous? Let a function $f \colon \mathbb{R} \to \mathbb{R}$ have the following two properties: 
(1) For all $x_1, x_2 \in \mathbb{R}$ such that $x_1 < x_2$, we have 
$$f \left( x_1 \right) \leq f \left( x_2 \right). $$
(2) For all $x_1, x_2 \in \mathbb{R}$,  we have 
$$f \left( x_1 + x_2 \right) = f \left( x_1 \right) + f \left( x_2 \right).  $$ 
Is such a function $f$ continuous at every point $c$ of $\mathbb{R}$? 
My Effort: 

We can show that for every rational number $q$, we have 
  $$ f(q) = q f(1). $$
As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable. 

How to proceed from here?
If we could show that $f$ is continuous at every rational point $c \in \mathbb{Q}$, then $f$ would also be continuous at every point of $\mathbb{R}$. Am I right?
What next? 
Context: 
Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition. 
 A: If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) \ge f(r_n) = r_n f(1) \to x f(1),$$
so we conclude $$f(x) \ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) \le xf(1)$. Hence, $$f(x) = xf(1) \quad \forall x\in\mathbb R.$$
(edit) I note the relevant portions of the book,



and,



it seems that $(a)$ you misread the book, which says you only need to prove continuity at one point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $\mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_n\to y$ then $x_n := y_n - y + x \to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) \to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
A: Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$.  Let $x_n \rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $\lim_{n \rightarrow \infty } f(x_n-c) = f(0)$. Hence $\lim_{n \rightarrow \infty } f(x_n-c) - f(-c)= \lim_{n \rightarrow \infty } f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
