My book gives proof as follows:
Thm
Let $I:= [a,b]$ be a closed bounded interval and lef $f:I\to \mathbb{R}$ be continuous on $I$. Then $f$ is bounded on $I$.
Proof
Suppose that $f$ is not bdd on $I$. Then, for any $n\in \mathbb{N}$ there is a number $x_n \in I$ such that $|f(x_n)| >n$.
I agree.
Since $I$ is bdd, the sequence $X = (x_n)$ is bdd. Therefore the Bolzano-Weierstrass Thm implies that there is a subsequence $X' = > (x_{n_r})$ of $X$ that converges to a number $x\in \mathbb{R}$.
Agreed.
Since $I$ is closed and the elements of $X'$ belong to $I$, it follows that $x\in I$. Then $f$ is continuous at $x$, so that $(f(x_{n_r}))$ converges to $f(x)$. And all convergent sequences are bdd.
Agreed. Note that this text uses " ( ) " for sequences rather than " { } ".
But this is a contradiction since $$|f(x_{n_r})|>n_r>r$$ Therefore the supposition that the continuous function $f$ is not bounded on the closed bounded interval $I$ leads to a contradiction. Q.E.D.
I just don't see the contradiction. My argument is that just because there exists a convergent sequence in $I$ making a specific sequence $(f(x_{n_r}))$ converge to a bdd $f(x)$ that doesn't mean that there doesn't exists some value on I where f(x) is unbdd. I think this proof would be fine if there was an added statement like "and every $x \in I$ is a cluster point of $I$" implying that we can generate a convergent sequence to each point. But i feel like there's something more I'm missing here or misinterpreting.
Since $x_{n_{r}}\rightarrow x$, then by the sequential characterisation of continuity of $f$, then $f(x_{n_{r}})\rightarrow f(x)$, so $|f(x_{n_{r}})|\rightarrow|f(x)|$. But the inequality $|f(x_{n_{r}})|>r$ forces that $|f(x_{n_{r}})|\rightarrow\infty$.