This is not a hard question but I do not know how to figure it out. Let $X$ and $Y$ be a nonempty sets. Let $\langle P,\leq\rangle=\langle P,\supset\rangle$ be be partially ordered where $P=\{f:D\to Y\colon D\subset X \ |D|<\omega\}$. If $|Y|\leq\omega$ then $P$ is ccc, ccc means every antichain at most countable.
Of course, if $P$ is countable, then it follows. Assume $P$ is uncountable, I think the way that can be done is , let $$ F=\{f_\alpha\colon A_\alpha\to Y\colon A_\alpha\subset X \text{ and } |A_\alpha|<\omega\}$$ and $F\subset P$ and $|F|\geq\omega_1$. It is enough to find two elements in $F$ that are compatible, that is, $f, g\in F$ and $ h\in F$ such that $f\subset h$ and $g\subset h.$ Also, $\Delta$-system lemma will be used to get $P$ ccc.
My attempt, let $f_\alpha, f_\xi\in F$ and show there are compatible. I think we can assume all $A_\alpha\neq A_\xi$ for all $\alpha, \xi<\omega_1$ Then by using $\Delta$-system there is a finite set $R$ such that $A_\alpha\cap A_\xi= R$.
If what I said was correct, how I can finish? Otherwise how this can done? Any help will be useful.
Let G be an uncountable subset of P and let $G\supset F=\{f_a: a\in \omega_1\}\subset P$ where $a\ne a'\implies f_a\ne f_{a'}.$ Let $dom(f_a)=A_a$.
Let $B$ be an uncountable subset of $\omega_1$ such that $D=\{A_a: a\in B\}$ is a $\Delta$-system. Let $A=\cap D.$
There are only countably many $f:A\to Y$ because $A$ is finite and $Y$ is countable. So $\psi(f_a)=f_a|_A$ maps the uncountable set $F^*=\{f_a:a\in B\}$ to the countable set $E=\{f_a|_A:a\in B\}.$
So there exists an uncountable $F^{**}\subset F^*$ and an $e\in E$ such that $\psi(f)=e$ for all $f\in F^{**}.$ Now take just $two$ unequal $f,f'\in F^{**}. $ We have dom($f_a)\cap $dom $(f_{a'})=A$ (because $D$ is a $\Delta$-system), and $f|_A=\psi (f)=e=\psi(f')=f'|_A,$ so $f,f'$ are compatible.
Remark. $f|_A$ means the restriction of a function $f$ to the domain $A$, when $A\subset $ dom($f).$