Here is my attempt at the proof.
Suppose that $a_n \rightarrow a$ and $b_n \rightarrow b$. Let $\varepsilon > 0$ be ABF and let $\varepsilon_1 = \frac{\varepsilon}{4}$ and $\varepsilon_2 = \frac{\varepsilon}{10}$. Then we know there exists $N_1, N_2 \in \mathbb{N}$ such that for all $n > N_1$ we have that $|a_n - a| < \varepsilon_1 = \frac{\varepsilon}{4}$ and for all $n > N_2$ we have that $|b_n - b| < \varepsilon_2 = \frac{\varepsilon}{10}$. Then $|2a_n - 2a| < 2\cdot\frac{\varepsilon}{4}$ and $|5b_n - 5b| < 5\cdot\frac{\varepsilon}{10}$. Let $N = \max\{N_1,N_2\}$. Then for all $n > N$ we have that
$$\begin{split} |c_n - (2a - 5b)| &= |(2a_n - 5b_n) - (2a - 5b)|\\ & = |(2a_n - 2a) - (5b_n - 5b)|\\ &\leq |2a_n - 2a| + |5b_n - 5b|\\ &< 2\cdot\frac{\varepsilon}{4} + 5\cdot\frac{\varepsilon}{10}\\ &= \varepsilon\\\ \end{split}$$
Is this proof correct?
Yes, you are completely correct.