Are lines in arbitrary normed vector spaces closed?

Question

Let $(V, \| \cdot \|)$ be a normed (real) vector space. Given two vectors $a$ and $d$ (with $d$ not the zero vector), is the line $ L = \{a + td: t \in \mathbb{R}\} $ through $a$ in direction $d$ necessarily a closed set?

In $\mathbb{R}^n$, this seems easy (for instance by showing that the line is polyhedral), but my different proofs all seem to rely on its finite-dimensionality. Making me suspect that the answer in general is probably no$\ldots$

Answer 1

To show a set is closed, you just have to take an arbitrary point not in that set and find an open neighborhood of that point which does not intersect your set. So, is there a point not on the line but such that every open neighborhood of the point meets the line? If so, what point?

Answer 2

Take any convergent sequence $(x_n)$of points on this line. Since each $x_n$ is on this line there is a sequence $(t_n)$ such that $x_n=a+t_nd$ where $0 \le t_n \le 1.$ Show that limit of $x_n$ is also on this line. Then you will have the closedness.