This is from Jay Cummings Real Analysis 1.34
Question: For each $n\in \mathbb{N}$, assume we are given a closed interval $I_n=[a_n,b_n]$. Also, assume that each $I_{n+1}$ is contained inside of $I_n$. Prove that $\bigcap^\infty_{n=1}I_n$ is non-empty.
Proof by induction:
(Base case) $n=1 \ a_1$ is contained within $I_n$, this is the only set in the interesction therefor it is non-empty.
$n=m+1$
$a_{m+1}$ is in the interval, by induction it is proved that $a_{m+1}$ is in every interval $I_b \ 0<b < m+1$.
The base case $b={m+1}$ is clear.
$b=j-1 \ \ \ \ \ \ $ $I_j\subseteq I_{j-1} \, \ \ \ \ \ \ \ a_{m+1}\in I_j\implies a_{m+1}\in I_{j-1}$
Therefor for $n=m+1$ it is true that $a_{m+1}$ is in $I_n$, therefor it is nonempty.