Prove $(C_{1} + C_{2})^{\perp}=C_{1}^{\perp} \cap C_{2}^{\perp}$ for any linear code $C_{1}, C_{2}$ over $\mathbb{F}_{q}$ of the same length.
we know $C^{\perp}= \{ x \in \mathbb{F}_{q}^{n}: <x,v>=0 \forall v \in C\}$, o $C^{\perp}=\{aH:a \in \mathbb{F}_{q}^{n-k}\}$ where $H$ the control matrix.
Let $x\in (C_1+C_2)^\perp$. Then for each $v\in C_1$ you have that
$\langle x,v\rangle=0$
so $x\in C_1^\perp$ while for each $w\in C^2$ you have
$\langle x,w\rangle =0$
so $x\in C_2^\perp$. This means that $x\in C_1^\perp\cap C_2^\perp$. Then
$(C_1+C_2)^\perp\subseteq C_1^\perp\cap C_2^\perp$
Let $x\in C_1^\perp\cap C_2^\perp$ and $w=v_1+v_2\in (C_1+C_2)$ where $v_1\in C_1$ and $v_2\in C_2$. Then
$\langle x,w\rangle =\langle x, v_1+v_2\rangle =$
$=\langle x,v_1\rangle +\langle x,v_2\rangle=0+0=0$
So $x\in (C_1+C_2)^\perp$ and this means
$C_1^\perp\cap C_2^\perp\subseteq (C_1+C_2)^\perp $