Use Nested Interval Theorem to prove that a continuous f is bounded on a closed interval $[a,b]$ 
Attached is my proof. Is there any flaw in my proof?
Thanks in advance.
I will use NIT to show a continuous $f$ is bounded on a closed interval $[a,b]$
I'll argue by contradiction.
Suppose f is not bounded on $[a,b]$
Then either 
a) $f(x)> N$ on $[a,\frac{a+b}{2}]$ or 
b) $f(x)> N$ on $[\frac{a+b}{2},b]$
let $I_0=[a_0,b_0]$
By reasoning, let $I_n=[a_n,b_n]$
then f(x)>N on whichever $[a_n,\frac{a_{n-1}+b_{n-1}}{2}]$ or $[\frac{a_{n-1}+b_{n-1}}{2},b_n]$
By NIT, there exists an $\alpha \in [a_n, b_n]$ for all $n \in N$
For $a_n, \alpha \in I_n$ $|a_n-\alpha|$ $<= \frac{b-a}{2^n}$ for all $n$
If $\epsilon>0 $ choose $N>n$ such that $\frac{b-a}{2^n}<\epsilon$
then $|a_n-\alpha|<\epsilon$
Hence $\lim_{x\to\infty}a_n=\alpha$ $\lim_{x\to\infty}f(a_n)=\alpha$
Similarly, $\lim_{x\to\infty}b_n=\alpha$ $\lim_{x\to\infty}f(b_n)=\alpha$
Hence f is bounded on $\alpha$, a contradiction
 A: The idea is not incorrect, but what you want to say is that if $f$ is not bounded on $[a,b]$, then it is not bounded in one of the halves. In particular, you want you $N$ to vary. Your proof is not clear as to how you're choosing this $N$ given the condition that $f$ is not bounded.
Since $f$ is not bounded in $[a,b]$, there exists a point in one half whose image is larger than say $1$ in absolute value. Now look at this half. Since $f$ is not bounded by our choice of half, there is one half of this interval where $f$ is not bounded, so there is a point there whose image is larger that $2$ in absolute value. Repeating this process, you can consutruct a sequence of points $x_1,x_2,\ldots$ and a sequence of nested closed intervals $I_1,I_2,\ldots$ such that $$\begin{align}(1)\hspace{.5cm}& x_i\in I_i\\ (2)\hspace{.5cm}& |f(x_i)|>i\\ (3)\hspace{.5cm}& {\rm length}(I_i)< |b-a|/2^i\end{align}$$
You can then choose $x\in \bigcap I_i$ and show that $f(x)$ cannot be finite, which is impossible. 
