What do metric spaces look like? I know that a metric is a way of measuring distance that can be generalized to sets outside of the real numbers.  I also know that the space is simply the set you are working in.  So is a metric space just the outcome of putting your set through the metric?
For example, if we have $(\mathbb{R}, d_0)$ with $d_0$ being the discrete metric does $\mathbb{R}$ simply become $\left\{0, 1\right\}$ because they are the only possible outcomes, like taking an open or closed ball?  More likely, a set is the same no matter what metric it is under, but then why can the same set be open in one metric and closed in another? (asides from induced metrics I kind of understand that)
I've done quite a lot of work with metrics at this point and still feel uneasy with them, especially with non-Euclidean metrics since the Euclidean metrics are the ones I've worked with the most.  I guess my question is what do spaces look like once they are under the effect of different metrics?  
I'm sorry this is a badly formatted and vague question, but I don't really know what I'm trying to ask.  I just can't get my head around the same space under different metrics and what they look like.  If anyone has examples or explanation that might give me a new understanding I would be very grateful.  If this question should be deleted, tell me and I will.
 A: A metric space is a formal object. It doesn't (and doesn't need to) "look like" anything. Trying to picture them can be useful, but most of times is of little importance. What is important is why you are considering a given metric. For example, if you want to measure how far away a real function on the interval is from the $0$ function, in some applications you want to consider that it is far if there is a value of $f$ which is far away. In others, you want to consider that it is is not very far if there is a small area underneath its graph.
Since it is a formal object, it is psychologically natural to try to grasp the concepts in terms of what you already understand. But that completely misses the point of the power of the abstraction, which is to surpass the confort zone of what you already understand.  
That being said, I think that there is an implicit mathematical question in your soft question. Maybe what would make you happy is among the following lines

Given a metric space $(X,d)$, is there a way to visualize it inside $\mathbb{R}^3$?

Which can be satisfactorily restated mathematically as

Given a metric space $(X,d)$, is there an isometry of $X$ with a subspace of $(\mathbb{R}^3,d_{\text{euclid}})$?

An isometry, if you are not familiar with the term, is a function which preserves distance.
For example, the abstract metric space $X=\{A,B,\text{Lettuce}\}$ with discrete metric admits an isometry to a subset $T$ of $\mathbb{R}^3$: take the vertices of an equilateral triangle in the plane.
However, not every metric space can be visualized like that. This is of course trivial if you consider a set of very big cardinality and put any metric you want on it (e.g., the discrete). However, even metrics on subsets of $\mathbb{R}^3$ itself may not be realized as a subspace of $\mathbb{R}^3$. For example, the same discrete metric in $\mathbb{R}$ cannot be visualized in $\mathbb{R}^3$-*.
Therefore, you can't reasonably imagine $\mathbb{R}$ with the discrete metric as a cluster of points in space in the way I mentioned. But that is not so harsh as it sounds: you don't need to. 
*For a proof, one way to see that is by noting that a subspace of a secound-countable topological space needs to be secound-countable. But the discrete metric on an uncountable set is not secound countable. This shows further that not even topologically it can be realized as a subset of Euclidean space.
A: Different metric on the same set sometimes induces different topology, you can think a metric on a set is like some sort of partition of the set into open sets.
You may think that different metric on the same set partitions the set differently.
