Suppose $A$ is a set. If $\mathcal{F}$ and $\mathcal{G}$ are partitions of $A$, then we’ll say that F refines G if $\forall X\in\mathcal{F}\exists Y\in\mathcal{G}(X\subseteq Y)$.
Let $P$ be the set of all partitions of
A, and let $R = \{(F, G) ∈ P × P | F$ refines $G\}.$
I am required to prove that $R$ is Anti-symmetric is the presented proof correct.
Proof. Assume that for some $\mathcal{F}\in P$ and $\mathcal{G}\in P$ it is the case that $(\mathcal{F},\mathcal{G})\in R$, $(\mathcal{G},\mathcal{F})\in R$ and $\mathcal{F\neq G}$. The proposition $\mathcal{F\neq G}$ suggests that $\mathcal{F}\Delta\mathcal{G}=(\mathcal{F\backslash G})\cup(\mathcal{G\backslash F})\neq\varnothing$, arguing from cases.
Case-1($\mathcal{(F\backslash G)}\neq\varnothing$): Assume that $\mathcal{F\backslash G}\neq\varnothing$ then it must be that there exists an $X$ such that $X\in\mathcal{F}$ and $X\not\in\mathcal{G}$ and since $\mathcal{F}$ refines $\mathcal{G}$ it follows that for some $Y\in\mathcal{G}$, $X\subseteq Y$ but $\mathcal{G}$ refines $\mathcal{F}$ and thus for some $Z\in\mathcal{F}$, $Y\subseteq Z$ summarizing we have $X\subseteq Y\subseteq Z$ evidently $X\subseteq Z$ and since $X\in\mathcal{F}$ consequently $X\cap Z=X\neq\varnothing$ and since $\mathcal{F}$ is pairwise-disjoint it follows that $X=Z$ but then our original deduction becomes $X\subseteq Y\subseteq X$ which implies that $X=Y$ and thus $X\in\mathcal{G}$ but we assumed that $X\not\in\mathcal{G}$ resulting in a contradiction.
Case-2($\mathcal{(G\backslash F)}\neq\varnothing$): We can arrive at a contradiction in a similar vein as in the above case.
$\blacksquare$
First, you need to break up that really long one sentence into several sentences!
Second, your proof is correct! Good job!
Third, it can be simplified a little bit. Notice that you end up showing that if $X \in \mathcal{F}$, then $X \in \mathcal{G}$ ... and of course you can likewise show that if $X \in \mathcal{G}$, then $X \in \mathcal{F}$. So, between these two results you can immediately conclude that $\mathcal{F}=\mathcal{G}$ without doing a proof by contradiction at all.