Quotient topology and normed vector spaces This is a question on an exercise in topology, more precisely in quotient spaces. 
Let $(E,||\cdot ||)$ be a normed vector space and $F\subseteq E$ a subspace. In the vector space $\widetilde{E}=E/F$ we define the map $p:\widetilde{E}\rightarrow\mathbb{R}$ with $p([x])=inf\{||x-y||:y\in F\}$. 
I am wondering if the quotient topology $\mathcal{T}_\pi$ on $\widetilde{E}$ is the same as the smallest (weak) topology $\mathcal{T}_p$ on $\widetilde{E}$ such that $p$ is continuous (that is, the sets $p^{-1}(G)$ such that $G$ is open in $\mathbb{R}$). 
In the notation, $\pi$ denotes the canonical projection $E\rightarrow\widetilde{E}$.

$p([x])$ is actually equal to $dist(x,F)$, the infimum of the distances of $x$ from the points of $F$.
Firstly, it is not hard to prove that $\mathcal{T}_p\subseteq \mathcal{T}_\pi$, using that $p\circ\pi$ is continuous. This is true because $E$ is also a metric space with the metric induced by the norm and in that case, $p\circ\pi$ is the usual $dist$ function on metric spaces and we know that it is continuous.
My question occurs when showing that $\mathcal{T}_\pi\subseteq \mathcal{T}_p$ holds. Let $U$ be open with the quotient topology, then we need to find an open subset $G$ of $\mathbb{R}$ such that $p^{-1}(G)=U$. This is the same as showing that $p$ is open under the quotient topology. 
Under what conditions does this hold?
Finally, the textbook mentions that if $F$ is a closed subspace, then $p$ is a norm on $\widetilde{E}$. Is it true that the topology induced by a norm is the same as the smallest topology such that the norm is continuous as a map? If so, why? We know that in metric spaces, norms are continuous maps.
 A: 
Is it true that the topology induced by a norm is the same as the smallest topology such that the norm is continuous as a map?

No, this is not true. This is not even true in dimension $1$: the topology induced on $\mathbb{R}$ by the absolute value $\mathbb{R} \to \mathbb{R}_{\geq 0}$ does not distinguish a point and its opposite since they are sent on the same point (their neighborhoods are exactly the same).
The topologies induced by $\pi$ and $p$ never coincide (except when $F$ is dense), for the same reason. But the topology induced by the seminorm $p$ is always the same as the topology induced by $\pi$.
Given a metric space $E$ and a function $f : E \to F$, we define the metric induced by $f$ on $F$ as the “smallest metric such that $f$ does not increase the distances” (this can be made exactly the same definition as the quotient topology, in just a different setting). Concretely, this metric is defined as $d(x,y) := d(f^{-1}(x),f^{-1}(y))$ or alternatively $d(x,y) = \inf\,\{d(x',y') \,|\, f(x')=x, f(y')=y\}$. Note that $d(x,y)$ can be infinite or zero even when $x \neq y$ (I think a name for it would be an $\infty$-semimetric space).
With that construction, $F$ becomes a metric space $(F,d_f)$. So it induces a topology $\tau(d_f)$ on $F$. But $f$, as a function from a topological space to a set, also induces directly a topology $\tau_f$. Then $\tau_f$ is always finer than $\tau(d_f)$, since $f : E \to (F,\tau(d_f))$ is continuous because it is $1$-lipschitz. In the case of a quotient of a normed vector space, the converse is also true and these two topologies are equal, but not in general.
To prove that $\tau_\pi \subseteq \tau(d_\pi)$, take an open set $O \in \tau_\pi$ and a point $x \in O$. The goal is to prove that $d_\pi(x,\complement O) > 0$. Choose $x' \in \pi^{-1}(x)$. The distance we want to compute is then
$$\begin{align*}
d(\pi^{-1}(x),\pi^{-1}(\complement O)) &= d(x'+F,\pi^{-1}(\complement O))\\
&= d(x',\pi^{-1}(\complement O)-F)\\
&= d(x',\pi^{-1}(\complement O))\\
&= d(x',\complement \pi^{-1}(O))\text{.}
\end{align*}$$
But this is greater than $0$ since $\pi^{-1}(O)$ is open and that $x' \in \pi^{-1}(O)$.
Moreover, the metric induced by $p$ is $d_\pi$.
A: On a metric space $(M, d)$, where $d: M\times M\to \mathbb{R}^+$ is the metric function, the $d$-induced topology is the smallest topology on $M$ such that the corresponding product topology of $M\times M$ contains the initial topology of metric function $d$. In fact, this is true when $d$ is only a psuedometric.
The essence of the norm topology on a normed vector space $(X, \|\cdot\|)$ can be understood along the same line of thinking. Note, however, the metric function $d$ is a bivariate function. If denoting $-: X\times X\to X$ as the vector subtraction function defined as $-: (x_1,x_2)\mapsto x_1 -x_2$, then the norm-induced metric function is the composition $d_{\|\cdot\|}:= \|\cdot\|\circ -$.
So the correct statement is: the norm topology $\tau_{\|\cdot\|}$ on $X$ is the smallest topology on $X$ such that the corresponding product topology on $X\times X$ contains the initial topology of metric $d_{\|\cdot\|}$.
The same is true when $\|\cdot\|$ is only a seminorm.
