One possible answer could be: by opening any book on vector bundles, and looking at the proposition right after the definition of stable vector bundle.

Another possible answer is as follows.

What you ask is valid in much more generality on any compact Kähler manifold.

**Proposition.** *Let $\mathcal F$ and $\mathcal G$ two torsion free coherent sheaves over a compact Kähler manifold $(X,\omega)$. Then, $\mathcal F\oplus\mathcal G$ is $\omega$-semistable if and only if $\mathcal F$ and $\mathcal G$ are both $\omega$-semistable with $\mu(\mathcal F)=\mu(\mathcal G)$.*

Here, $\mu(\mathcal E):=\deg_\omega\mathcal E/\operatorname{rank}\mathcal E$, where
$$
\deg_\omega\mathcal E:=\int_X c_1(\mathcal E)\wedge\omega^{n-1},\quad\dim X=n,
$$
for any torsion free coherent sheaf $\mathcal E$.

*Proof.* Suppose first that $\mathcal F$ and $\mathcal G$ are both $\omega$-semistable with the same slope $\mu$. Then, $\mu(\mathcal F\oplus\mathcal G)=\mu$. Given a subsheaf $\mathcal E$ of $\mathcal F\oplus\mathcal G$, set $\mathcal E_1=\mathcal E\cap(\mathcal F\oplus 0)$ and $\mathcal E_2$ to be the image of $\mathcal E$ under the projection $\mathcal F\oplus\mathcal G\to\mathcal G$. By the semistability of $\mathcal F$ and $\mathcal G$ you have
$$
\deg_\omega(\mathcal E_i)\le\mu\cdot\operatorname{rank}\mathcal E_i,\quad i=1,2.
$$
Therefore,
$$
\mu(\mathcal E)=\frac{\deg_\omega(\mathcal E_1)+\deg_\omega(\mathcal E_2)}{\operatorname{rank}\mathcal E_1+\operatorname{rank}\mathcal E_2}\le\mu,
$$
and $\mathcal F\oplus\mathcal G$ is $\omega$-semistable.

Conversely, since both $\mathcal F$ and $\mathcal G$ are at the same time quotient sheaves and subsheaves of $\mathcal F\oplus\mathcal G$, you have 
$$
\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).
$$  
But any subsheaf $\mathcal E$ of $\mathcal F$ or $\mathcal G$ is a subsheaf of their direct sum, as well. Hence,
$$
\mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square
$$