In asymptotic notation how to prove that $\mathcal{O}(g(n))\subseteq\mathcal{O}(f(n))\implies\mathcal{O}(f(n)+g(n))=\mathcal{O}(f(n))$
I think you are pretty much done.

$$
g(n) \ge 0 \implies f(n) \le f(n) + g(n), \quad \forall n \in \mathbb{N}
$$

So pick $c = 1 \in \mathbb{R}^+$ and $n_0 = 1 \in \mathbb{N}$, then
$$
f(n) \le c\cdot(f(n) + g(n)), \quad \forall n\ge n_0
$$