Logic: unique function; chain

Logic: unique function; chain

Suppose $X$ and $Y$ are sets. Let $P$ be the set of all pairs $(A,f)$
where $A$ is a subset of $X$ and $f$ is a function $A \rightarrow Y$. Then
$P$ is a poset with the following relation: $(A,f) \leq (B,g)$ iff $A
\subseteq B$ and $f$ is the restriction of $g$ to $A$.
Show that, if $C= \{ (A_i,f_i) |i \in I \}$ is a chain in $P$, there is a
unique function $f: \cup_{i \in I}$ $A_i \rightarrow Y$ such that for each
$i$, $f_i$ is the restriction of $f$ to $A_i$.
I've got some vague ideas to solve this problem, such as the use of
induction on the chain (to obtain the function in different steps of
expansion) and a proof by contradiction for the uniqueness of the function
(if it was not unique, then it already was in some step of the induction,
which gives us a contradiction). But somehow i'm struggling to make a
correct mathematical proof out of this.
Any help would be welcome and much appreciated!