Definition of topology..."open in $X$" does it not involve a metric by definition? I've been very confused with topology lately, and here's another one that I can't answer myself after spending hours staring at the definition.
So, here's the definition I am using

$X$ is a set. A topology on X is a set of subsets $\tau$ of $X$ with the following properties

*

*Whenever $(U_{i})_{i \in I}$ is a family(finite or not) of subsets of $X$ such that $U_i \in \tau$, $\forall i \in I$ then $\cup_{i \in I}U_i \in \tau$


*Whenever $U_1$, $U_2 \in \tau$ then $U_1 \cap U_2 \in \tau$.


*$\emptyset \in \tau$ and $X \in \tau$

So far so good. But Here's a remark in my notes that confuses me;

We call the elements of $\tau$ the open subsets of $X$. Thus "$U \in \tau$" and "$U$ is open in $X$" mean the same thing.

Firstly, "open subsets", in my understanding, is that "$U$ is an open subset of $X$, iff $\exists$ open ball $B_\epsilon(X)$ of radius $\epsilon >0,  \forall x \in U \subseteq X$ such that $B_\epsilon(X) \subseteq X$"
Casually speaking, for every point in this subset, I can find an open ball around it which is always in that subset.
But to define an open ball, we need a metric, by definition of a ball, yes?
If the elements of $\tau$ are called the "open subsets" of $X$, then we clearly are talking about the existence of a ball, thus the existence of a metric. But just as in the definition, $X$ is a set, but no where does it say it is a "metric space"
Is it always assumed that $X$ has some metric with it and we're building a "topology" over it? If so, then why do we have something called a "topology induced by a metric"? If we're always assuming metrics defined on $X$ to create a topology(because, without it, we cannot define "open" thus there would be no such thing as "open subsets"), doesn't that mean every possible topology must be induced by some metric?? I am so confused
And, for the remark, it's saying that being a member of $\tau$ is the same as being called an "open subset" and "being open in $X$".
But, why? The definition does not mention anything about $U_i$ being open or closed or whatever,can I not take closed subsets of $X$ and still satisfy the $3$ conditions i.e. qualify them as a topology?
Please someone help me out here, every time I feel like I understood something, a new question comes up and I feel depressed...Every time I get something, it contradicts with my understanding...
 A: This is something that also caused me a lot of troubles when I started topology. The key to understanding what is going on at first is not to think about what you learned in calculus about open sets. In topology, an open set is just an element of your topology $\tau$, and $\tau$ must satisfy the axioms you mentioned. Therefore, being open is something that is totally dependent on which topology you're considering and is not an intrisic property of subsets of $X$. 
If $X$ is a metric space, you can define $\tau$ to be the set of subsets $U\subset X$ that satisfy the "ball condition". It is a good exercise to show that $\tau$ is indeed a topology. In that case, when you work with this topology, being open in the topological sense is the same as being open in the sense of calculus.  But you could also consider other topologies on $X$ that don't have anything to do with the metric and you would get totally different open sets. 
So you just have to think of the topology $\tau$ as some extra structure on $X$ telling you which subsets you can call open. 
A: The issue is that we don't need a metric to study continuity, because $f\colon X \to Y$ is continuous if and only if for every $U \subseteq Y$ open, $f^{-1}(U)\subseteq X$ is open (here $X$ and $Y$ are topological spaces). We also know that an arbitrary union of open sets is again open, same for finite intersections. Also, $X$ and $\varnothing$ are open.
We try to "axiomatize" that, saying that a topology $\tau$ is a collection of sets in $X$ verifying the conditions in the definition you have there.
Then we say that the sets in $\tau$ are "open", and as far as we go here, "open" is nothing more than a name.
You can check that if you have a metric space $(X,d)$, then the collection $\tau_d$ of "$d$-open" sets (defined with open balls, using the metric $d$) do verify conditions 1., 2. and 3. of the definition of topology.
A: Not every topology is induced by a metric.  For example, let $X$ be an infinite set and $\mathscr T=\{U\subseteq X:X\setminus U\text{ is finite or } U=\emptyset\}$.  You can easily check that $\mathscr T$ satisfies the definition of a topology.  So, any set whose complement is finite is considered to be open in this topology.
A: What you've learned previously is that given a metric space, $X$, we can define a topology on $X$.
Note that it is quite possible for two different metrics on $X$ to define the same topology on $X$. Since continuity under a metric only depends on the topology, this means that the exact metric is much less important than the open sets that come from the metric. (Metrics have other purposes, such as defining uniform continuity, something which does not exist in general topology.)
Also, there are topologies on $X$ that do not come from any metric. 
For example, if $X=\mathbb R$, we can define a topology I'll call $\Lambda$, for "left," where the open sets are of the form $(-\infty,\alpha)$, for $\alpha\in\mathbb R$, and the empty set and all of $\mathbb R$. This is a perfectly fine topology by the above definition, but it does not correspond to a metric.
A: There is interesting spaces where the topology cannot be given by a metric one.
A simple exemple would be the spaces of functions from $\Bbb R$ to $\Bbb R$ for the pointwise convergence (or, in other words, the product topology on $\Bbb{R}^\Bbb{R}$)
