Let $X$ be a compact interval, let $V$ be a normed vector space and suppose that a map $f:X \rightarrow V$ is unbounded. I'm trying to see why there must exists a sequence $(x_i)$ in $X$ such that $|f(x_i)| \geq i \; \; \forall i \in \mathbb{N}$.
My argument is as follows: Since $f$ is unbounded, $sup\{|f(x)| \; : x \in X\}$ does not exist and therefore $1$ is not a supremum. This means that there is some $x_1 \in X$ such that $f(x_1) \geq 1$. We can select $x_2, x_3$ and so in in a similar manner and by induction conclude that there exists a sequence $(x_n) \in X$ such that $|f(x_i)| \geq i, \; \; i \in \mathbb{N}$
My questions are:
(1) I didn't use compactness and I don't believe that its a necessary hypothesis; is this correct? and
(2) This argument selects an infinite number of elements of $X$, more-or-less simultaneously, which I believe actually requires the axiom of choice. Does the induction principle preclude the need to use the choice axiom here or should it be applied to make the argument correct?