Using Winther result I can finish the proof for other cases. In order to do this we have to deduce this problem to rationals.

According to little bit stricter version of Dirichlet's approximation theorem we can find infinitely many pairs of fraction $\left(\frac{p_1}q, \frac{p_2}q\right)$ with the following property:

$$\max\left(\left|a - \frac{p_1}q\right|, \left|b - \frac{p_2}q\right|\right) < q^{-\frac32}.$$

Infinitely many means that we can make $q$ as large as we want, we will choose it later. Finally, let's take $n = q-1$:

$$
\{n\frac{p_1}{q}\} + \{n\frac{p_2}{q}\} = \{1-\frac{p_1}{q}\} + \{1-\frac{p_2}{q}\} = 2 - \frac{p_1}{q}-\frac{p_2}{q} =\\
= 2 - \{c\} + \left(a - \frac{p_1}{q}\right) + \left(b - \frac{p_2}{q}\right) = 2 - \{c\}+\frac{\alpha}{q}, \alpha \in [-2, 2]
$$

Also from Dirichlet's theorem we know that $n\left(a - \frac{p_1}{q}\right) < q^{-\frac12}$ and the same thing with $b$. Finally we can get:

$$
2 - \alpha q^{-\frac12} = \{nc\}+\{c\} \leq 1 + \{c\}, \alpha \in [-4, 4]
$$

Now we can take very large $q$ and get contradiction.

**Edit**. In fact I omitted one little issue, namely, why the following implication is true?

$$
(na - n\frac{p_1}q) < q^{-\frac12} \Rightarrow (\{na\} - \{n\frac{p_1}q\}) < q^{-\frac12}
$$

Unfortunately, it is not true in every single case, but in our situation we can handle it. Remember that $n = q-1$. The problem can occur if there is an $A \in \mathbb{Z}$ (in fact $A = p_1$) such that

$$
n\frac{p_1}q=(q-1)\frac{p_1}q = p_1 - \frac{p_1}q < A < (q-1)a = na
$$

And also we know that distance between $n\frac{p_1}q$ and $na$ less than $q^{-\frac12}$. Last thing that we should mention is that $\frac{p_1}q > \frac{a}2 > 0$ since $\frac{p_1}q$ is close to $a \neq 0$.