Question On the Proof of The boundedness Theorem let $f:[a,b]\rightarrow\mathbb{R}$, f continuous on $[a,b]$. I shall prove that $\exists A,B\in\mathbb{R}, \forall x\in[a,b], A\le f(x)\le B$.
Proof: Let's define $g(x)=|f(x)|$, we need to prove now that $\exists A\in\mathbb{R}, \forall x\in[a,b], g(x)\le A$. let's suppose that this claim is false, therefore we get:
$\forall A, \exists x\in[a,b], g(x)> A$.
$ \forall n\in\mathbb{N}$ we define $A=n$ therefore $\exists x_n\in[a,b], g(x_n)>n$.
We know that $a\le x_n \le b$ and from B.W. we know that $\exists x_{n_k}$ that converges to $y\in\mathbb{R}$. furthermore, we also know that $a\le y \le b$.
$g(x_n)>n$ therefore $g(x_{n_k})\ge n_k$
Because $g$ is continuous on $[a,b]$ w know that $\lim_{k\to \infty}g(x_{n_k})=g(y)\mathbb\in{R}$.
We also know that $\lim_{k\to \infty}n_k=\infty$.
Up to this point I understand each sub-claim in the proof. However, at this point in the proof it's deduced that we can compare the limits and say that because the first sequence converges to a number in $\mathbb{R}$ and the second converges to $\infty$ and there is a $ \ge $ sign between them there is a contradiction. 
Could you please make this comparison more formal and explain why I can deduce this? To be more clear: Why when given two sequences that converge I can apply $\ge$ on their limits and what does it mean when a sequence that converges to a number in $\mathbb{R}$ is greater or small than $\infty$.
Thanks.

Your analysis is fine.
You have constructed a sequence so that $g(x_{n_k}) \rightarrow \infty$, which contradicts the fact that $g(x_{n_k}) \rightarrow g(y)$. It is a contradiction because a sequence cannot converge to two limits (or diverge properly to $\infty$ and converge to a real number at the same time).
