Given sequence of nested intervals $I_n[a_n,b_n]$. Show that if $y=\inf\{b_{1},b_{2},b_{3},\ldots\}$ then $y\in[a_{n},b_{n}]$ $\forall n$ Let $I_{1}=[a_{1},b_{1}]$, $I_{2}=[a_{2},b_{2}]$, $I_{3}=[a_{3},b_{3}]$,
$\ldots$ be a sequence of closed bounded nested nonempty intervals
$I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq\cdots$. Show that if
$y=\inf\{b_{1},b_{2},b_{3},\ldots\}$ then $y\in[a_{n},b_{n}]$ $\forall n$
Since the set $\{b_{1},b_{2},b_{3},\ldots\}$ is bounded below by $a_{1}$, by the axiom of completeness the $\inf\{b_{1},b_{2},b_{3},\ldots\}$ exists let then let $y=\inf\{b_{1},b_{2},b_{3},\ldots\}$. Fix any $n$ we will show that $y\in[a_{n},b_{n}]$ $\forall n$. By the definition of $y$ being the infimum we know that $y\leq b_{n}$, $\forall b_{n}$. Thus if we can show that $a_{n}$ is a lowerbound for  $\{b_{1},b_{2},b_{3},\ldots\}$ then $a_{n} \leq y$ where $y$ is the greatest lower bound. We will show that $a_{n} \leq b_{k}$, $\forall k$. 
Case $1$: If $k \geq n$ then $a_{k} \leq a_{n} \leq b_{k} \leq b_{n}$ by the nested interval property. 
Case $2$: if $k<n$ then we have that $a_{n} \leq a_{k} \leq b_{n} \leq b_{k}$ by the nested interval property. 
Hence we have shown that $a_{n} \leq b_{k}$, $\forall k$ thus $y\in[a_{n},b_{n}]$ $\forall n$.

You have made slight errors in the inequalities in both the cases.
If $k\geq n$, then the nested property gives $[a_n, b_n]\supseteq [a_k, b_k]$ so $a_n\leq a_k\leq b_k$ and therefore $a_n\leq b_k$.
If $k<n$, then $[a_k, b_k]\supseteq [a_n, b_n]$ so $a_n\leq b_n\leq b_k$ and so $a_n\leq b_k$ again.
