Can a standard topological space be constructed from a closed subset in $\mathbb{R}^n$? I have just started studying topological spaces. I am having trouble understanding the relationships between closed spaces. My question is:

Can a standard toplogical space be constructed from $X$, a closed subset in $\mathbb{R}^n$, which includes all points in $X$?

My Understanding
Definition of a Topological Space: A topological space is an ordered pair $(X, T)$, where $X$ is a set and $T$ is a collection of subsets of $X$ which satisfy the following condition:


*

*$\emptyset,X\in T$ and $T$ is closed under arbitrary unions and finite intersections.  


I understand that the standard topology on $\mathbb{R}^n$ is $(\mathbb{R}^n,T)$, where $T$ is the set of all possible unions of open balls in $\mathbb{R}^n$. 
Similarly for the open set $X\subset \mathbb{R}^n$, a standard topology can be given by $(X,T_X)$, where $T_A$ is the set of all possible unions of open balls in $X$.
I do not understand the case where $X$ is a closed subset. Can I use the set of all closed subsets in $X$? Can I only use open sets to construct a topological space?
 A: The construction that you're looking for is called the subspace topology. Given a topological space $(Y, \tau)$, and set $A \subset Y$, the topology on $Y$ induces a topology $\tau_A$ on $A$ called the subspace topology of $Y$ on $A$ by the rule $$ V \in \tau_A \text{ if and only if } V = A \cap U \text{ for some } U \in \tau.$$ Regardless of the topological space $(Y, \tau)$ and subset $A$, this is in fact a topology. Sets $U \in \tau_A$ are called relatively open in $A$ with respect to $Y$, or just open in $A$ when it is understood that the subspace topology is being used. This is the standard way of introducing a topology on a subset of a space, and works without problem on $\mathbb{R}^n$ with the standard topology.
A: Here is an attempt to answer part of your question:

I do not understand the case where $X$ is a closed subset. Can I use the set of all closed subsets in $X$? Can I only use open sets to construct a topological space?

I'm not exactly sure what you mean by "use the set of all closed subsets": You can use them to tell you which sets are open; a set is "open" iff its complement is closed. So, if you have a list of all closed subsets of $X$, then the open sets in your topological space would be precisely the complements of these closed subsets in $X$.
You don't want to "use" closed subsets in the sense of forming a topological space as articulated in your definition. I'll come back to this point at the end.$^\star$
As an example, suppose $X = [0,1] \subset \mathbb{R}$ is the space in question. Here, the open sets would be exactly those that can be written in the form $U \cap X$ where $U \subset \mathbb{R}$ is open in the traditional sense.
Example: The set $(1/3, 2/3)$ is open in $X=[0,1]$ because we can use this set itself as the above-mentioned $U$ and observe that $U \cap X = U$. So $U = (1/3, 2/3)$ is open in $X$. Similarly, any open set in $\mathbb{R}$ that is contained entirely in $X$ is also open in $X$.
But there are other sets that are open with this "subspace topology" that were not open in $\mathbb{R}$.
Example: The set $[0, 1/2)$ is open in $X = [0,1]$. Note that this set is not open (nor is it closed, incidentally) in $\mathbb{R}$. But, it is open in our topological space whose set is $X$ with the subspace topology: specifically, we can write the set $[0, 1/2)$ as the intersection $(-1, 1/2) \cap X$. Since the set $(-1, 1/2)$ is open in $\mathbb{R}$, we have that $[0, 1/2) = (-1, 1/2) \cap X$ is open in $X$. (But not $\mathbb{R}$!)
$^\star$: You cannot just take closed subsets and declare them the open sets. For example, if you naively tried to use all closed subsets of $[0,1]$ to form a topology, you'd run into problems. For example, one of your requirements for a topological space is closure under arbitrary unions. But, the set formed by the union of closed sets 
$$\bigcup_{n>1} [\frac{1}{n},\frac{n}{n+1}] = (0,1)$$
yields an open set, not a closed one; so, the idea of using only closed sets fails: it is not closed under arbitrary unions as demonstrated by this example in which an arbitrary union of closed sets yields a set that is not closed.
