Does a Banach space always contain an element of arbitrarily large norm?

Question

Let $X$ be a Banach space. Or say, even just a normed linear space. Let $N \in \mathbb N$.

Does there exist $x \in X $ with $\|x\| \ge N$?

If $X$ is a Banach space then its unit ball is also a Banach space so the answer should be no. But in the proof of the Alaoglu theorem it is used that if we let $D_x = B(0,\|x\|) \subseteq \mathbb C$ then $\mathbb C = \bigcup_{x \in X} D_x$ and this seems to me requires that for every $N$ there is $x$ with $\|x\|\ge N$.

Answer 1

Let $x\neq 0$ be any non-zero element, so that $\lVert x \rVert\neq 0$. Let $y=\dfrac{N}{\lVert x \rVert}x$. Then $\lVert y \rVert = N$. Right?

Answer 2

If $X \neq \{0\}$ is a normed space:

Take any $x \in X \setminus \{0\}$. For any $N \in \Bbb N$, $\frac{N}{\|x\|}x$ has norm $N$. Remember that by definition, normed spaces are vector spaces.