Is $\mathbb Q$ a metric space? 1) Is $\mathbb Q$ a metrice space ? I would say yes since I would take $|x-y|$ as a norm for all $x,y\in\mathbb Q$, but I don't see to what would look the topology induced by $|\cdot |$, is it the discrete topology ?
2)Other question, is $\mathbb Q$ a $\mathbb Q$ vector space ? I would say yes, but the it should be a banach space since it's dimension is finite, but a banach space is complete whereas $\mathbb Q$ is not since there is a sequence in $\mathbb Q$ that converge for example to $\pi$. So what the correct answer ?
 A: You mix up different notions: a metric on some set $X$ is a function $d:X\times X\rightarrow\mathbb{R}$ satisfying


*

*$\forall x,y\in X\quad d(x,y)\geq 0$,

*$d(x,y)=0\Leftrightarrow x=y$,

*$\forall x,y,z\in X\quad d(x,z)\leq d(x,y)+d(y,z)$.


In the case of the rationals $d(x,y):=|y-x|$ is indeed a metric. The induced topolgy is not discrete, because a set $U\subseteq\mathbb{Q}$ is open, if and only if for every point $x\in U$ there exists an "open ball"
$B(x,r)=\{y\in\mathbb{Q} : d(x,y)<r\}$
contained in $U$. Such an open ball always contains infinitely many rationals.
As for your 2nd question: a finite dimensional vector space over the reals is always complete but not over the rationals.
A: Questions 1):
Yes, $\mathbb{Q}$ equipped with $d(x,y) = |x-y|$ is a metric space. The reason is that $\mathbb{R}$ equipped with $d(x,y) = |x-y|$ is a metric space, and when you have any metric space $X$ with metric $d$ and any subset $Y \subset X$, the restriction of $d$ to $Y$ defines a metric on $Y$. Furthermore, the metric topology on $Y$ equals the subspace topology on $Y$ induced from the metric topology on $X$.
No, the topology on $\mathbb{Q}$ equipped with the metric $d(x,y) = |x-y|$ is not the discrete topology. It is instead the subspace topology inherited from $\mathbb{R}$. This topology has a basis given by sets of the form $\mathbb{Q} \cap (x-\epsilon,x+\epsilon)$ for any $x \in \mathbb{R}$ and any $\epsilon > 0 \in \mathbb{R}$. A single point $p \in \mathbb{Q}$ does not form an open set in this topology on $\mathbb{Q}$, which would be a good exercise to verify.
As an aside, as it is stated your question "Is $\mathbb{Q}$ a metric space?" has a minor flaw, because it is incomplete without putting the proposed metric into the question. It would have been better to ask the question in the form "Is $\mathbb{Q}$ equipped $d(x,y) = |x-y|$ a metric space?" The reason this is important is that there are many other candidates for metrics $d(x,y)$ which you may encounter in your mathematical life, most importantly the $p$-adic metrics.
Questions 2):
Here, your question "Is $\mathbb{Q}$ a vector space?" is incomplete without putting the proposed ground field into the question. This is really a serious flaw in the question, which led you to misunderstand the relation to Banach spaces. So the question as stated has no answer.
If instead you had asked "Is $\mathbb{Q}$ a vector space over the field $\mathbb{Q}$?" then the answer is "yes". 
Or if you had asked "Is $\mathbb{Q}$ a vector space over the field $\mathbb{R}$?" then the answer is "no".
