Open in X (X is a metric space) how can you show or tell if it is open or not in general terms? What does it really mean? I suppose analysis and abstract math isn't really my cup of tea... So, I am looking at "open" and "closed" but here I want some more explanation on "open".
I am now on a course in Topology and being rusty with these fundamental definitions just makes me so wobbly.
Say, my notes says;

It is not really proper to say "open" sets; properly, it should be, if X is a metric space, then a subset U of X is "open in X."
Open in X means that, iff for $\forall u \in U, \exists \epsilon>0$ such that $B_\epsilon(u) \subseteq U$ where $B_\epsilon(u)$ is the open ball.

Okay, while this is hard to visualize say, in $n$ dimensions higher than $3$, I think I understand it. My problem is, I can't really apply it. Say the following very basic statement;

$[0,1)$ is open in $[0,2]$ but not open in $\mathbb{R}$.

Why? I guess the ultimate reason is "You cannot define an open ball in the latter for any $\epsilon>0$" but why? How do you know? How can you tell at a glance? I mean, all I know is that $X$ is a metric space but I am not told what the metric is; How can you say that no open ball exists for "any" possible metric you can dream of on the interval and $\mathbb{R}$?
Maybe I just feel like I understand but really don't about the "open  in X" thing.
It might be great if someone who had some hard time understanding these concepts explain to me how they got over it. Some people get these things in 2 seconds and don't seem to quite understand people like me who gets confused.
Thanks so much for the help!
 A: Perhaps what's confusing you is the common way of referring to the metric space $X$ without mentioning the actual metric; but it's always understood to be known. 
Whenever you have a question about whether something is or is not an open, closed etc. set in a space $X$, do not ever think of it in terms "how do I know, maybe for some metric..." This is a tell-tale sign that you're confused. The metric is fixed and known; the question refers to this fixed and known metric and not to anything else. It just often goes unmentioned explicitly, that's all, to avoid being tedious.
So to take your question:

$[0,1)$ is open in $[0,2]$ but not open in $R$.

There are two metric spaces in this question. One is $R$ which is tacitly understood to carry its usual metric $d(x,y) = |x-y|$. You know this metric well. The other is $[0,2]$ considered as a space on its own, which is tacitly understood to "inherit" the metric from $R$. So the metric there is the same, and also known to you. 
Now to the actual substance of the question. How is it possible that the metric is the same, the set $[0,1)$ belongs to both spaces, but it's open and one and not open in the other? How do you tell at a glance?
"Open" means "every point of the set comes with a little ball around it that's also entirely in the set". Always use this intuition, preferably geometrically (thinking in pictures). How do we tell at a glance if $U$ is an open set? We look at a point of $U$ and check whether we can draw a ball around it that stays in $U$. (Often it'll be enough to imagine doing this for just one "typical" point of $U$. But not in this example. Here the point $0$ behaves differently from others). Then you translate this geometric intuition into a formal proof if needed.
So... in $R$, the set $[0,1)$ is not open, why? Because every ball around $0$ (and in $R$, a ball around $x$ is some interval $(x-\epsilon, x+\epsilon)$ must include a negative point, and so is not entirely in $[0,1)$. 
In $[0,2]$, why is the same set open? Because there an open ball around $0$ - a set of all points with distance $< \epsilon$ from $0$ - is an interval $[0,\epsilon]$ that's entirely in $[0,2]$. What changed? The negative numbers are no longer in the space, we took them away, so they're not part of balls in the space $[0,2]$. The space is a subspace of $R$, it has the same metric, but because some points are missing, the balls can be different.
A: Let $A$ be a subset of a metric space $X$.  A subset $U$ of $A$ is called open in $A$ iff there's an open subset $V$ in $X$ such that $U=V\cap A$. Hence $U=[0,1)$ is open in $A=[0,2]$ because $V=(-1,1)$ is open in $X=\mathbb R$ and $U=A\cap V$,i.e., $[0,1)=[0,2]\cap(-1,1)$.
