Boundedness theorem, why the limit of subsequence lies in the interval at which the original sequence is bounded?

Question

Theorem: if $f$ is continuous on closed interval $I$ then it is also bounded. The proof is given by contradiction, assuming that for any $n$ it is possible to find $x_n$ such that it is greater than n $f(x_n)>n$. It is proven to be impossible by noting that because $x_n$ is bounded sequence it has a subsequence $x_{n_k}$ that is convergent and the limit again lies within the interval $I$. Why is it so? The second question is about the contradiction where it is stated that because the subsequence is convergent it is also bounded which is in contradiction to a fact that $|f(x_{n_k})|>n_k>k \\ \forall k \in N$. I don't understand why $n_k \ge k$?

Answer 1

Since the interval $I$ is closed, when a sequence $(x_n)_{n\in\mathbb N}$ of elements of $I$ converges, its limit belongs to $I$.

And whenever we have a subsequence $(x_{n_k})_{k\in\mathbb N}$ of a sequence $(x_n)_{n\in\mathbb N}$, the map $k\mapsto n_k$ is strictly increasing and therefore $n_1\geqslant1$, $n_2>n_1\implies n_2\geqslant 2$, $n_3>n_2\implies n_3\geqslant3$, and so on…