To prove an apparently obvious statement: if $A_1subseteq A_2 subseteq ... subseteq A_n$, then $bigcup_{i=1}^n A_i = A_n$

Question

To prove:$$text{If } A_1subseteq A_2 subseteq ... subseteq A_ntext{ ,then } bigcup_{i=1}^n A_i = A_n$$ using the axioms of ZFC Set Theory.

Honestly, this statement is very obvious, but I do not want to take that for granted. How can I prove this from the axioms, especially the axiom of unions?
I'm fairly new to ZFC Set Theory, so I don't know where to start. I do know that the Axiom of Unions (and Specification) gives rise to the definition of $bigcup_{i=1}^n A_i$ for sets $A_1,...,A_n$ though. The uniqueness follows from the Axiom of Extensionality.
My thoughts:
We already know that the union exists, and it is defined such that for every element $xin bigcup_{i=1}^n A_i$, there exists at least one set $A_j$ such that $xin A_j$. What's next?
Thank you!
Edit:
Induction works if the chain is finite or countably infinite. What if the chain length is uncountable? Does this result still hold, and if yes, how do we prove it?
Edit 2:
So now I wish to prove (or disprove) the following using the axioms of ZFC Set Theory:

Consider an infinite set $A$, and an index set $I$ such that $$A_1subseteq A_2subseteq ... subseteq A_i subseteq ... subseteq A$$
where $iin I$. The index set $I$ is uncountable. We have $$bigcup_{iin I}A_i = A$$

Infinity_hunter · Accepted Answer

This is regarding your second edit. First of all denoting $A_1 subseteq A_2 .... subseteq A$ is not correct when the index set $I$ is uncountable and moreover you don't know the elements of $I$.
Take $A_i =(0,i)$ where $iin (0,1)$ and $A = [0,1]$. These are nested in the sense that $i leq j$ implies that $A_i subseteq A_j$ and observe that for $i$ in index set $A_i subset A$. Clearly union of all $A_i$ is the interval $(0,1)$ which is not $A$.

Luiz Cordeiro · Answer

To be completely formal, you'd could first prove by induction that $A_isubseteq A_n$ for every $ileq n$. Then prove that $bigcup_{i=1}^n A_isubseteq A_n$ and that $A_nsubseteqbigcup_{i=1}^n A_i$. This last one should be trivial by definition of the union.
As for the first one, if $xin bigcup_{i=1}^nA_i$, then $xin A_j$ for some $j$, and then we know (from the proof by induction) that $xin A_n$.

To prove an apparently obvious statement: if $A_1subseteq A_2 subseteq ... subseteq A_n$, then $bigcup_{i=1}^n A_i = A_n$

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