Let $X$ an affine space and $S,T$ two affine subspace of $X$. I want your opinion on my proof about the statement 
If $S \cap T = \emptyset$ its okay. If $S \cap T \neq \emptyset$, we have $P \in S \cap T$. Let $P_S \in S$ and $P_T \in T$, then by Chasles $$ \overrightarrow{P_SP_T} = \overrightarrow{P_SP} + \overrightarrow{PP_T} \in \overrightarrow{S} + \overrightarrow{T} $$ But $S,T$ parallel so $\overrightarrow{S} = \overrightarrow{T}$. In addition $\overrightarrow{T},\overrightarrow{S}$ are vector spaces, then we got : $$ \forall P_S \in S \mbox{, } \forall P_T \in T \mbox{, } \overrightarrow{P_SP_T} \in \overrightarrow{S} + \overrightarrow{T} = \overrightarrow{T} = \overrightarrow{S} $$ Now, let a point $Q \in S$, then $$ \forall P_T \in T \mbox{, } \overrightarrow{QP_T} \in \overrightarrow{T} \Rightarrow Q \in T $$ So, $S \subset T$. Now, let a point $Q \in T$, then the same strategy leads to $Q \in S$. Conclusion $S = T$.