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, $$ \mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square $$

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, you have $$ \mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G). $$
Indeed one has the following lemma, which follows almost immediately from the definition of semistability.

Lemma. If $$ 0\to\mathcal S\to\mathcal E\to\mathcal Q\to 0 $$ is a short exact sequence of coherent sheaves on a compact Kähler manifold $(X,\omega)$, then $$ \operatorname{rank}(\mathcal S)(\mu(\mathcal E)-\mu(\mathcal S))+\operatorname{rank}(\mathcal Q)(\mu(\mathcal E)-\mu(\mathcal Q))=0. $$

Now, this lemma permits to define semistability also in terms of quotient sheaves. Namely, $\mathcal E$ is semistable if and only if for every quotient sheaf $\mathcal E\to\mathcal Q\to 0$ of positive rank one has $\mu(\mathcal E)\le\mu(\mathcal Q)$.

But then, since both $\mathcal F$ and $\mathcal G$ are at the same time quotient sheaves and subsheaves of $\mathcal F\oplus\mathcal G$, the equality of slopes follows.

Finally, any subsheaf $\mathcal E$ of $\mathcal F$ or $\mathcal G$ is a subsheaf of their direct sum, as well. Hence, $$ \mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square $$

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, $$ \mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square $$

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, $$ \mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square $$