My proof goes like this:
By definition
$f \in \{ g:\mathbb{N} \rightarrow \mathbb{R}|\exists \beta > 0 \exists n_0 \in \mathbb{N} \forall n \geq n_0 : 0 \leq f(n) \leq \beta g(n) \}$
therefore $0 \leq \beta_f g(n) \leq f(n) \,\forall n \geq n_f$ .
Let $h \in \mathcal{O}(g):=\{f:\mathbb{N} \rightarrow \mathbb{R}|\exists \alpha > 0 \exists n_0 \in \mathbb{N} \forall n \geq n_0 : 0 \leq f(n) \leq \alpha g(n)\}$
and therefore $0 \leq h(n) \leq \alpha_h g(n)\,\forall n \geq n_h$.
Rearrange the inequality to $\frac{h(n)}{\alpha_h} \leq g(n)$.
Substitute g(n) into $0 \leq \beta_f g(n) \leq f(n) \,\forall n \geq n_f$.
Resulting to $0 \leq \frac{\beta_f}{\alpha_h} h(n) \leq f(n) \,\forall n \geq max \{ n_{h}, n_{f}\}$
Rearrange inequality: $0 \leq h(n) \leq \frac{\alpha_h}{\beta_f} f(n)$
This means $h \in \mathcal{O}(f)$ and therefore $\mathcal{O}(g) \subseteq \mathcal{O}(f)$.
Thanks in advance.