I'd like a proof verification. I tend to forget/miss details. I hope it is right, if not I'd love to here what went wrong.
Also, if you know another way to show that statement, I'd be very curious to hear about it.
Statement:
Let $I_n=[a_n,b_n]$ be a sequence of intervals such that $\forall n\in\mathbb{N}, I_{n+1}\subset I_n$. Show that $\bigcap_{n=0}^{+\infty}I_n\neq\emptyset$
Proof:
If $I_{n+1}\subset I_n$, then the sequence $(I_n)_{n\in\mathbb{N}}$ is decreassing. We notice that $I_{n+1}\subset I_n\iff[a_{n+1},b_{n+1}]\subset[a_n,b_n]$, which implies that $(a_n)_{n\in\mathbb{N}}$ is increassing and $(b_n)_{n\in\mathbb{N}}$ decreasing. Because $\mathbb{R}$ is an ordered set, $a_n\leq b_n$, which implies that $\exists(a,b)\in(\mathbb{R}\cap I_0)^2$ such that $a_n\leq a$ and $b_n\geq b,\forall n\in\mathbb{N}$. As $(a_n)_{n\in\mathbb{N}}$ (resp. $(b_n)_{n\in\mathbb{N}}$) is increasing (resp. decreasing) and has an upperbound (resp. lowerbound), both $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ converge and $\bigcap_{n=0}^{+\infty}I_n=[a,b]$, where $a=$sup$\{a_n|n\in\mathbb{N}\}$ and $b=$inf$\{b_n|n\in\mathbb{N}\}$.
Also, if lim$_{n\rightarrow +\infty}(a_n-b_n)=0$, then $a=b$
Best Answer
One nitpick I can see is that you inferred the intersection is $[a,b]$ by considering limits of the interval endpoints. However, it's not obvious that $a,b\in \cap_{n=0}^\infty I_n$ from the point of view of sets. It becomes clearer if you state that since $[a_{n+1},b_{n+1}]\subseteq [a_n,b_n]$, if $a\notin \cap_nI_n$, then $a\notin I_{k}$ for some $k>0$, which implies either $a<a_k$ or $a>b_k$, both of which will yield contradictions.
As an alternative, you can try to show that $[a,b]\subseteq \cap_{n=0}^\infty I_n$ and $[a,b]\supseteq \cap_{n=0}^\infty I_n$.