Dual space & orthogonal $F_1^{\perp}+F_2^{\perp}\subset(F_1 \cap F_2)^{\perp}$ 
$E$ is a finite-dimensional vector space over a field $K$ and $E^∗$ its dual.
$F_1$ and $F_2$ is a subspace of $E$
I would like to prove that $F_1^{\perp}+F_2^{\perp}\subset(F_1 \cap F_2)^{\perp}$

I asked this question 12 hours ago, but with no answser and no comment, I've prefered to cancel the last one and to rewrite it correctly.
Is my proof correct?

Let $F_1^{\perp}:=\bigg\{\gamma\in E^*, \quad \forall x\in F_1, \;\gamma(x)=0\bigg\}$ and $F_2^{\perp}:=\bigg\{\psi\in E^*, \quad \forall x\in F_2, \;\psi(x)=0\bigg\}$
$\varphi \in F_1^{\perp}+F_2^{\perp},\iff\exists (\gamma,\psi)\in \left(F_1^{\perp}\times F_2^{\perp}\right), \forall x \in E \quad \varphi(x)=(\gamma+\psi)(x)$
We can deduce as well $\exists \varphi'\in F_1^{\perp}+F_2^{\perp}$ such that $\varphi'(x)=(\gamma-\psi)(x)$
Now let : $F :=\bigg\{x\in E\;| \quad \forall \phi \in (F_1^{\perp}+F_2^{\perp}), \;\phi(x)=0\bigg\}$ and any $\varphi \in F_1^{\perp}+F_2^{\perp}$
Thus
$\forall x\in F, \left\lbrace\begin{array}{l} \varphi(x)=0\\ \varphi'(x)=0\end{array}\right. \iff \left\lbrace\begin{array}{l} \gamma(x)+\psi(x)=0\\\gamma(x)-\psi(x)=0\end{array}\right.\iff \left\lbrace\begin{array}{l} 2\gamma(x)=0 \quad \color {red}{(1)}\\2\psi(x)=0 \quad \color {red}{(2)}\end{array}\right. $
Since $(1)\land (2)$ we deduce that $ \forall \varphi \in F_1^{\perp}+F_2^{\perp},\;\forall x\in F,\quad \varphi(x)=0 \implies x\in F_1\cap F_2$
We can conclude $\varphi \in (F_1\cap F_2)^{\perp}\quad \square$

You can shorten this a lot. There are a great deal of extraneous constructions that miss the main idea. Since you originally reference a dual space, I assume you are asking about dual spaces, not orthogonal complements and have answered accordingly; be careful, as they are not the same thing.
Let $\phi\in F_1^0+F_2^0$, where $S^0 = \{\rho\in E^* \ | \ \forall x\in S, \  \rho(x)=0 \}$ denotes the annihilator of $S$. (It looks like this is perhaps what you mean by your $S^\perp$ notation.) Then $\phi = \gamma + \psi \ $  for some $\gamma\in F_1^0, \psi\in F_2^0$. Now if $x\in F_1\cap F_2$, then $x\in F_1$ and $x\in F_2$, so
$$\phi(x) = (\gamma + \psi)(x) =\gamma(x) + \psi(x) = 0 + 0 = 0.$$
Hence, $\phi\in(F_1\cap F_2)^0$.
