There is not enough space...!? I read at Wikipedia about space:

I understand that


*

*a metric space is a set for which elements a metric is definied.

*the Euclidean space uses the Euclidean metric

*a Hilberts space generalizes the Euclidean space to spaces of infinite dimensionality. I guess, this means it also uses the Euclidean distance.

*a Banach space does not necessarily use the Euclidean metric as a metric.


Q1: What do you call a space, that is like the Euclidean space in every regard, except it uses a different metric - let's say the Manhatten metric? I guess that would make this space a multivariate real coordinate space ($R^n$) that uses a different metric?
Q2: Where did the metric definition "get lost"? In other words, what do you call a space that is like a Hilbert space in every regard, except it uses a different metric - let's say the Manhatten metric?
Shouldn't the graphic than look more like this:

 A: First let me address your question "Shouldn't the graphic look like this?".
If you follow the link you provided back to that Wikipedia page, and if you read the caption below that diagram, and if you also check each arrow to verify, you will see that the diagram has the following feature: each arrow $A \to B$ can be interpreted as "Every type $A$ object is an example of a type $B$ object". Thus, for example, finite dimensional Euclidean space is an example of a Hilbert space, which is an example of a Banach space, which is an example of a Normed vector space, and so on.
Your diagram does not have that feature. Suppose you take something in your Q2 box, which (I think) means any metric on $\mathbb R^n$. The result might not be an inner product space, nor even a Banach space. I could, for example, pick a diffeomorphism $f : \mathbb R^2 \to B^2$, where $B^2$ denotes the open ball, and then I could pull the $B^2$ metric back to $\mathbb R^2$ using the function $f$. The resulting metric on $\mathbb R^2$ would have finite diameter. But, every Banach space, and every inner product space, has infinite diameter. 
Regarding your Q1, $\mathbb R^2$ equipped with the Manhattan metric is an example of a Banach space. In fact, any norm on any finite dimensional vector space (such as $\mathbb R^n$) defines a metric that is automatically complete, and therefore it is an example of a Banach space. (Similarly, any finite dimensional vector space, equipped with a metric realized by a dot product, is necessarily complete, and therefore is a Hilbert space.)
Regarding your Q2, you have given two versions of it, one version in your diagram, and a different version in the comments. Let me address both. Q2 in your diagram is handled by my earlier example using a diffeomorphism between $\mathbb R^n$ and an open ball in Euclidean $n$ space. Q2 in your comment is handled by an important fact: the Manhattan metric is not realized by any dot product, so it is a Banach space, and it is not a Hilbert space. 
