Prove that $E^*\subseteq F^{\bot}\oplus G^{\bot}$. 
Let $F$ and $G$ be closed subspaces of the Banach space $E$ such that
$E=F\oplus G$. How can I show that $E^*\subseteq F^{\bot}\oplus
G^{\bot}$, where $E^*$ is the topological dual of $E$ and
$$F^{\bot}=\{\varphi\in E^*:\varphi(x)=0, \text{ for all }x\in
 F\}\text{ }\text{ }\text{ and }$$
$$G^{\bot}=\{\varphi\in
 E^*:\varphi(x)=0, \text{ for all }x\in G\}?$$

My attempt was:
Take $\varphi\in E'$, arbitrary. So, for all $x\in E$, $x=x_f+x_g$, with $x_f\in F$ and $x_g\in G$. So, $$\varphi(x)=\varphi(x_f+x_g)=\varphi(x_f)+\varphi(x_g).$$
As $E=F\oplus G$, so $F\cap G=\{0\}$. And I think that information will guide me to the conclusion. But how can I show that $\varphi\in F^{\bot}\oplus G^{\bot}$?
 A: We have $E = F \oplus G$. Let $\pi_F: E\to F$ and $\pi_G: E\to G$ denote the canonical projection maps. These are well-defined, since $F\oplus G$ is a direct sum, i.e. for every $x\in E$, there exist unique $x_f\in F, x_g\in g$ with $x = x_f + x_g$.
For every $\varphi \in E^*$, we have the (unique) decomposition
$$\varphi = \varphi \circ \pi_F + \varphi \circ \pi_G $$
where $\varphi \circ \pi_F \in F^\perp$ and $\varphi \circ \pi_G \in G^\perp$.
From here, you may conclude that $E^* \subset F^\perp \oplus G^\perp$. I leave the details to you.
A: Following your idea, you can write $\phi=\phi_F+\phi_G$, where $\phi_F(x_f+x_g):=\phi(x_f)$ and $\phi_G(x_f+g_g):=\phi(x_g)$.
What you have to show then is , that $\phi_F$ and  $\phi_G$ are  continous.
$\phi_F$  for example can be decomposed as
$$
\phi_F:F\oplus G\overset{\mathrm{pr}}\to F \overset{\mathrm{inc}}\to F\oplus G\overset{\phi}\to \mathbb K
$$
The crucial point is the continuity of the first projection map. Here you have to use, that $F$ and $G$ are closed and that $E$ is banach, see for example the answer here.
