Let $(V,\mathcal{T})$ be a topological vector space where $\mathcal{T}$ is the topology induced by a norm
$$\Vert \cdot \Vert: V \to [0, \infty[$$
Is it true that $\mathcal{T}$ is the initial topology w.r.t. this norm?
Let $\mathcal{S}$ be the initial topology generated by this norm. I can see that $\mathcal{S}\subseteq \mathcal{T}$ must hold but does the other inclusion also hold?
EDIT: the other inclusion also seems to hold:
Consider the ball $B_{\Vert \cdot \Vert}(0,\epsilon)$. Since this is the inverse image of the open set $[0, \epsilon[$ under the norm map, we see that $B_{\Vert \cdot \Vert}(0, \epsilon) \in \mathcal{S}$. Since $(V, \mathcal{S})$ is also a topological vector space, all translates of this ball are also in $\mathcal{S}$. Thus $\mathcal{S}$ contains a basis of $\mathcal{T}$, and we must have $\mathcal{T}\subseteq \mathcal{S}$ as well.
Is this correct?
Given the norm function $n(x)=\|x\|: V \to \Bbb R_0^+$, it's certainly true that $\mathcal{T}$, the induced topology by that norm on $V$, does make $n$ continuous, so for $\mathcal{T}_n$ (the initial topology induced by $n$ on $V$) we can surely say $$\mathcal{T}_n \subseteq \mathcal{T}$$ by minimality.
The reverse is certainly not true: Because we have an initial topology induced by a single function, $$\mathcal{T}_n = \{n^{-1}[O]: O \subseteq \Bbb R^+_0 \text{ open}\}$$ and this implies that if $n(x)=n(x')$ for $x,x' \in V$ and any open $O \in \mathcal{T}_n$: $x \in O \iff x' \in O$, so $V$ in $\mathcal{T}_n$ is not $T_0$ and so quite different from the metric topology $\mathcal{T}$.
However, if we use the notion of a weak vector-space topology, things change: the minimal vector space topology $\mathcal{T}_{n,v}$ (making the $+,-,\cdot$ operations continuous) that also makes $n$ continuous, has the property that all norm-open balls are open in $\mathcal{T}_{n,v}$ as you state and then the fact that all operations are continuous allows you to translate those to all other points as well. So trivially yes to your title question, if we work in the category of TVS's and trivially no if we work in the category Top.