Equivalent norm induces equivalent metric 
Question:
Equivalent norms induces equivalent metric and equivalent metrics induces the same topology.

I have the definition on hand for what it means for metric, norms and normed space to be equivalent. What is unclear is the language "induced" used.
Any clarification to shed light is appreciated.
Thanks in advance.
Edit: 
Attempt:
Let V be a vector space over the field F of reals $\mathbb{R}$.
All norms $\left \| \cdot  \right \|: V\rightarrow \mathbb{F}$
satisfies the homogeneity condition
$\left \| \alpha \vec{v} \right \|=\left | \alpha \right |\cdot \left \| \vec{v} \right \|: \forall \alpha \in \mathbb{F}, \vec{v} \in V$
Define the metric 
$d: V \times V\rightarrow \mathbb{F}
\left ( \vec{x},\vec{y} \right )\mapsto \left \| \vec{x}-\vec{y} \right \|$ 
So the metric defined by the Norm is such that
$d\left ( \alpha \vec{x} ,\alpha \vec{y} \right )=\left | \alpha \right |\cdot \left \| \vec{x}-\vec{y} \right \|=\left | \alpha \right |d\left ( \vec{x},\vec{y} \right )$
Suppose $\left \| \cdot  \right \|, \left | \cdot  \right |$ are both equivalent Norms.
Then, we have
$\alpha_{1}d\left ( \vec{x},\vec{y} \right ) \leq e\left ( \vec{x},\vec{y} \right ) \leq \alpha_{2}d\left ( \vec{x},\vec{y} \right )$

Recall: A topology induced by a metric.
  Let $\left ( X,d \right )$ be a metric space and $\tau$ the collection of arbitrary union of open balls in X.
  $\tau$ is called the metric topology induced by the metric d.

By definition, this is the metric topology.
In the metric space $\left ( V,\left \| \cdot  \right \| \right )$ over the field $\mathbb{F} of reals \mathbb{R}$, the open balls are 
$B_{\epsilon }\left ( \vec{a} \right )=\left \{ \vec{x} \in V : d\left ( \vec{x},\vec{a} \right )  = \left | \vec{x}-\vec{a} < \epsilon \right |\right \} $
for $\epsilon >0, \vec{a} \in V$.
$\tau$ is a topology on V called the standard topology.
 A: Define the metric induced by a norm on a vector space $V$ as $$ d(x,y) = || x - y||. $$ Consider the case of the real numbers with the norm as the absolute value of a number. We indeed have the metric as the absolute difference in this  case.
The topology induced by a metric is the topology generated of all open balls in that space, i.e for any point in the space $X$, the neighborhood of radius $\varepsilon$ around that point is an open set. Again, compare with the standard topology on the reals with the usual metric.
A: The topology induced (generated) by a metric or a norm is the topology for which the the set of open balls is a base. Two metrics (norms) are equivalent iff they induce (generate) the same topology .
Metrics $d,e$ are called uniformly equivalent iff  there exists positive $k_1,k_2$ such that $k_1d(x,y)\leq e(x,y)\leq k_2d(x,y).$ Uniformly equivalent metrics are equivalent metrics, but two metrics can be equivalent without being uniformly equivalent.
On the other hand if two norms $\|\cdot \|_1$ and  $\|\cdot \|_2$ on a real or complex vector space are equivalent they  must be uniformly equivalent, in the sense that the metrics $d_1(x,y)=\|x-y\|_1$ and $ d_2(x,y)=\|x-y\|_2$ are unformly equivalent metrics. 
The standard topology on $\mathbb R$ is also called the usual topology on $\mathbb R.$
