Image of continuous $f:[a,b]\rightarrow \mathbb R$ is bounded Here is a proof of the theorem in title I'm working through.
This is a proof by contradiction, first assuming that it isn't bounded. Here I got the result that there is a sequence $\langle x_n \rangle \subseteq [a,b]$ such that $\forall \epsilon \in \mathbb R_{>0}, \exists n$ with $|f(x_n)|>\epsilon$.
The second result I got is that $\langle x_n \rangle$ has a subsequence $\langle x_{n_r}\rangle$ converging to a $\xi \in [a,b]$, which in turn implies $\displaystyle \lim_{r\rightarrow \infty} f(x_{n_r})=f(\xi)$.
The site I linked to says that these two results are contradictory, which I can't see - where is the logical contradiction between these two?
 A: 
there is a sequence $\langle x_n \rangle \subseteq [a\,..b]$ such that $\forall \epsilon \in \mathbb R_{>0}:\exists f(x_n):|f(x_n)|>\epsilon$

For simplicity, let's make a small, subtle change and let $x_n$ instead be a sequence such that $|f(x_n)|$ diverges to infinity (it's a subsequence of the one in the highlighted text).  In other words, $x_n$ is a sequence such that, for any $\varepsilon > 0$, we have $|f(x_n)| > \varepsilon$ for all sufficiently large $n$.
$\large{ \dagger}$ Next, let's set $\varepsilon >> |f(\xi)|$.  The above means that there exists an $N \in \mathbb{N}$ such that, for all $n \geq N$, we must have $f(x_n) > \varepsilon$. 

$\langle x_n \rangle$ has a subsequence $\langle x_{n_r}\rangle$ converging to a $\xi \in [a\,..b]$, which in turn implies $\displaystyle \lim_{r\rightarrow \infty} f(x_{n_r})=f(\xi)$.

Yes, this is a consequence of Bolzano-Weierstrass.  Now taking everything above together, this is the contradiction:
If $x_{n_r}$ converges to $\xi$, this means an infinite number of terms in $x_n$ must be "close" to $\xi$, which means we must have an infinite number of evaluations $f(x_{n})$ "close" to $f(\xi)$.  However, due to $\large{\dagger}$ above, only a finite number of terms can be "close" to $f(\xi)$ since $f(x_n) >> f(\xi)$ for all $n \geq N$.

There are some other methods of tackling this proof as well that are worth mentioning:
Alternative Method $1$:  One can prove that a continuous function on a compact set is uniformly continuous.  Recall that uniform continuity means, given an $\varepsilon > 0$ the same $\delta$ works at every $x$ in the domain.  Therefore, if we compute $\displaystyle n = \Bigg\lceil \frac{|b-a|}{\delta} \Bigg\rceil$, then the "diameter" of $f([a,b])$ cannot be larger than $\varepsilon \cdot n$.
Alternative Method $2$:  One can prove that the continuous image of a compact set is itself compact.  Since $[a,b]$ is compact, then so too is $f([a,b])$.  What are the necessary and sufficient conditions for a subset of $\mathbb{R}^n$ to be compact?
A: You are right, technically. 
The statement is ofc true, you just need to change notation a bit to make it clear… and a little bit more work to do.
You get the existence of a sequence $\langle x_n \rangle \subseteq [a\,..b]$ with $\forall \epsilon \in \mathbb R_{>0}:\exists f(x_n):|f(x_n)|>\epsilon$
So you can conclude there exists a subsequence $\langle x_{n_r} \rangle$ which diverges to $+\infty$ monotonously!
For this subsequence $\langle x_{n_r} \rangle\subseteq [a\,..b]$ you know there exists a subsubsequence $\langle x_{n_{r_q}} \rangle$ converging to $\xi$… and then you have your contradiction
A: The quoted proof uses a much stronger fact about the sequence $\langle x_n\rangle$, namely that
$$
\lim_{n\to\infty}|f(x_n)|=\infty
$$
In particular, for every subsequence,
$$
\lim_{r\to\infty}|f(x_{n_r})|=\infty
$$
which contradicts
$$
\lim_{r\to\infty}f(x_{n_r})=f(\xi)
$$
The existence of the sequence can be proved as follows: since $f$ is unbounded, for every $n$ there exists $x_n$ with $|f(x_n)|>n$.

A different strategy not using Bolzano-Weierstraß is by bisection. Set $a_0=a$ and $b_0=b$. Since $f$ is unbounded, it will be either in $[a_0..(a_0+b_0)/2]$ or in $[(a_0+b_0)/2..b_0]$ (or both). Choose the half-interval $[a_1..b_1]$ where $f$ is unbounded.
Such a construction can be repeated on $[a_1..b_1]$ and so on, leading to a nested family of intervals $[a_n..b_n]$, of width $(b-a)/2^n$, over each of which $f$ is unbounded.
It's easy to see that $\sup_n a_n=\inf_n b_n$, call this point $c$. Since $f$ is continuous at $c$, it is bounded in an interval $(c-\delta..c+\delta)\cap[a..b]$, for some $\delta>0$.
Take $n$ such that $(b-a)/2^n<\delta$ and you have a contradiction.
