Can a metric linear space induce a normed linear space? If $(X,\|\cdot\|,+,\cdot,F)$ is a normed linear space with addition $+$ and scalar multiple $\cdot$ over $F$. Then we can simply define a metric, called induced norm(?), $d:X\times X\to\mathbb{R}$ by $d(x,y)=\|x-y\|$ for all $x,y\in X$. Then $(X,d)$ become a metric space(or metric linear space).
However, conversely speaking, if $(Y,d')$ is a metric space, and if it is also possble to define the things needed so that $(Y,d',+,\cdot,F)$ become a metric linear space. Can we simply defined a norm function such that $(Y,\cdots\text{something}\cdots)$ become a normed linear space? In $\mathbb{R^n}$, we can simply do this. We can define $\|\mathbf{x}\|=d(\mathbf{x},\mathbf{0})$ for all $\mathbf{x}\in\mathbb{R}^n$, where $d$ is the euclidean metric, that is $d(\mathbf{x},\mathbf{y})=\sqrt{(x_1-y_1)^2+\cdots\cdots}$ for all $\mathbf{x}=(x_1,x_2,\cdots,x_n)$, $\mathbf{y}=(y_1,y_2,\cdots,y_n)$. But in a general metric space, is it always possible?
And this question arise the other question: which is the better(in some sense) order to define euclidean norm and euclidean metric in $\mathbb{R^n}$, pedagogically?

I don't think I fully understand the question, but it is not true that every metrisable topological linear space there is a norm which induces the topology. For instance, consider the discrete metric on $\Bbb R^n$. Even if one requires that the metric is translation invariant.
