This follows from the fact that retractions are type-preserving, because they are defined using isomorphisms between apartments. (See the argument used in the proof of (3.16) in Suzuki's book.)

In particular, in the case $m=0$ where $E \subset C$, you already know that $\phi(\phi'(C)) = C$, so $\phi(\phi'(E)) \subset C$. The only subset of $C$ of the same type as $E$ is $E$ itself, so you are done.

This follows from the fact that retractions are type-preserving, because they are defined using type-preserving isomorphisms between apartments. (See the argument used in the proof of (3.16) in Suzuki's book.)

In particular, in the case $m=0$ where $E \subset C$, you already know that $\phi(\phi'(C)) = C$, so $\phi(\phi'(E)) \subset C$. The only subset of $C$ of the same type as $E$ is $E$ itself, so you are done.

Edit:

Without using types, this seems to require a bit more detailed work. We can spell out $\phi(\phi'(E))$ as $$ \rho(\Sigma, C') \rho(\Sigma', C) \rho(\Sigma, C) \rho(\Sigma'', C') (E) , $$ and now notice that the product of the middle two retractions $\rho(\Sigma', C) \circ \rho(\Sigma, C)$ acts as the identity on $\Sigma'$. However, $\rho(\Sigma'', C') (E)$ is contained in $C''$ and hence in $\Sigma'$, so we get $$ \phi(\phi'(E)) = \rho(\Sigma, C') \rho(\Sigma'', C') (E) . $$ We now apply the same idea again: the product $\rho(\Sigma, C') \circ \rho(\Sigma'', C')$ is the identity on $\Sigma$, and we conclude that indeed $\phi(\phi'(E)) = E$.