I have already read this answer to the question Union of two affine subspaces, but I have tried to write down a different proof of the following proposition.
Let $\mathcal{E}$ be an affine space over a $K$-vector space $E$, where $\mathrm{char}\ K \neq 2$. Let's suppose $\mathcal{F}_1$ and $\mathcal{F}_2$ are affine subspaces of $\mathcal{E}$, whose directions are the vector subspaces $F_1$ and $F_2$ of $E$, respectively. The union $\mathcal{F_1 \cup F_2}$ is an affine subspace of $\mathcal{E}$ if and only if $\mathcal{F}_1 \subseteq \mathcal{F}_2$ or $\mathcal{F}_2 \subseteq \mathcal{F}_1$.
The ''if'' part is straightforward: if $\mathcal{F}_1 \subseteq \mathcal{F}_2$, then $\mathcal{F}_1 \cup \mathcal{F}_2 = \mathcal{F}_2$, which is an affine subspace by hypothesis; if $\mathcal{F}_2 \subseteq \mathcal{F}_1$, then $\mathcal{F}_1 \cup \mathcal{F}_2 = \mathcal{F}_1$, which is an affine subspace by hypothesis.
In order to prove the ''only if'' part, let's suppose $\mathcal{F}_1 \cup \mathcal{F}_2$ is an affine subspace, whose direction is the vector subspace $F$ of $E$. Assume the thesis is false. Then we can pick $A \in \mathcal{F}_1 \setminus \mathcal{F}_2$ and $B \in \mathcal{F}_2 \setminus \mathcal{F}_1$.
Since $\mathcal{F}_1 \cup \mathcal{F}_2$ is an affine subspace, the vector $\vec{AB}$ is an element of $F$ and there exists $P \in \mathcal{F}_1 \cup \mathcal{F}_2$ such that \begin{equation*} \vec{AP} = \frac{1}{2} \vec{AB}. \end{equation*}
Such a point $P$ must belong to $\mathcal{F}_1$ or to $\mathcal{F}_2$.
If $P \in \mathcal{F}_1$, then $\vec{AP} \in F_1$, and we also have $\vec{AB} \in F_1$ (because $\vec{AB} = 2 \vec{AP}$), which leads to the contradiction $B \in \mathcal{F}_1$.
If $P \in \mathcal{F}_2$, then $\vec{BP} \in F_2$, and we also have $\vec{BA} \in F_2$ (because $\vec{BA} = 2 \vec{BP}$), which leads to the contradiction $A \in \mathcal{F}_1$, and the proof is complete.
Is this proof correct?
It is correct.
You should however introduce $F$ as the vector subspace of directions of your putative affine subspace $\mathcal{F}_1\cup\mathcal{F}_2$ first. You also could introduce $\mathcal{F}:=\mathcal{F}_1\cup\mathcal{F}_2$ to make notation a bit lighter.