What does neighborhood/ball/closure mean in a non-metric and/or finite space? I'm trying to understand these fundamental concepts of topology. I understand what open closed and boundary mean in a continuous space with a metric, but I'm having issues understanding the meanings in cases that are finite or without a metric. 
I need a specific example to understand this, so lets use the following as an example:
$X=[1,4]$;
$Y =[1,4] \cap \mathbb{Z}$;
 $\mathscr{T}_e$ is the topology generated by the euclidean metric. 
$\mathscr{T}_c$={X,$\varnothing$[1,2), [1,3), [1,4)}
 $\mathscr{T}_d$={Y,$\varnothing$, {1},{1,2},{1,2,3}}
So we have 4 topological spaces: $(X,\mathscr{T}_e),(Y,\mathscr{T}_e), (X,\mathscr{T}_c), (Y,\mathscr{T}_d)$

Question 1: What are the neighborhoods of {3} in these spaces?

$(X,\mathscr{T}_e)$: I know here this would mean balls with a positive radius, centered at 3.
$(Y,\mathscr{T}_e)$: this would also be balls: {3}, {2,3,4}, {1,2,3,4}
$(X,\mathscr{T}_c)$: I'm not sure. Maybe all sets that contain 3, so  {1,2,3} would be the only neighborhood.
 $(Y,\mathscr{T}_d)$: Not sure either. Maybe [1,4) is only neighborhood?

Question 2: What is the meaning of a ball?

Let: $U=B_{1}(2)$
$(X,\mathscr{T}_e)$:I know here $U=$ (1,3)
$(Y,\mathscr{T}_e)$: Here $U=$  {2}
$(X,\mathscr{T}_c)$: not sure what a ball is here
 $(Y,\mathscr{T}_d)$:  not sure what a ball is here

Question 3: What is the meaning of closure?

If: $U=B_{1}(2)$, then what is $\overline{U}$?
$(X,\mathscr{T}_e)$:I know here $\overline{U}=$ [1,3]
$(Y,\mathscr{T}_e)$: Here i'm not sure either $\overline{U}$={2} or $\overline{U}$={1,2,3}
$(X,\mathscr{T}_c)$: not sure what a closure is here
 $(Y,\mathscr{T}_d)$:  not sure what a closure is here
 A: Ok, the first thing you have to understand is that open (closed) balls only exists in metric spaces. A distance function is needed to define them. In general topological spaces, openness(closedness) is defined entirely in terms of certain families of sets and their relations.
The other thing you need to understand is that a niehborhood of a point in the modern definition is not necessarily open, it merely has an open subset that contains the point. Of course,any open subset that contains the point is a nbhd of it, but not all nbhds are open. Let's see if we can clarify these definitions using the examples you gave.  
On question 1, you made an error-you mixed up the spaces (Y,$T_d$) and (Y,$T_c$).Let's look at what you wrote and I corrected.
*$X=[1,4]$;
$Y =[1,4] \cap \mathbb{Z}$;
 $\mathscr{T}_e$ is the topology generated by the euclidean metric. 
$\mathscr{T}_c$={X,$\varnothing$[1,2), [1,3), [1,4)}
 $\mathscr{T}_d$={Y,$\varnothing$, {1},{1,2},{1,2,3}}
So we have 4 topological spaces: $(X,\mathscr{T}_e),(Y,\mathscr{T}_e), (X,\mathscr{T}_c), (Y,\mathscr{T}_d)$

Question 1: What are the neighborhoods of {3} in these spaces?
  $(X,\mathscr{T}_e)$: I know here this would mean balls with a positive radius, centered at 3.
  $(Y,\mathscr{T}_e)$: this would also be balls: {3}, {2,3,4}, {1,2,3,4}
  : $(Y,\mathscr{T}_d)$I'm not sure. Maybe all sets that contain 3, so  {1,2,3} would be the only neighborhood.
   $(X,\mathscr{T}_c)$: Not sure either. Maybe [1,4) is only neighborhood?*

With my correction, we'll now not only check if they're correct,but understand why. Let's precisely define a general open set, an open ball centered at a point and a neighborhood of a point. 
Def: Let X be a nonempty set and let F be a family of subsets of X such that if G $\subseteq$ F: 
1)X,$\emptyset \in $ F
2) $\cup G \in$ F  
3) $\cap^n _{i\in I}G_i\in $ F  
Then F is called the topology on the set and any set in F is called an open set in X. The ordered pair (X,F) is called a topological space. A neighborhood U of a point x$\in$ X is the following set: x $\in A \subseteq U$ where A$\in$ F.   
These definitions hold for any topology on a nonempty set whether or not a metric is present.Note also that a nieghborhood doesn't have to be itself open. While we're at it, let's define a metric space: 
Def: Let X be a nonempty set and let d: X x X $\rightarrow \mathbb R$ be a mapping such that for any x,y,z $\in$ X :
 1) d(x,y) $\geq 0$ 
2) d (x,y) = 0 iff x = y. 
3) d(x,y) = d(y,x) 
4) d(x,y) $\leq$ d(x,z) + d(z,y)    
Then we say d is a metric on X and the ordered pair (X,d) is called a metric space. An open ball of radius $\epsilon$ $\geq 0 \in \mathbb R$ centered at x is the following set:
$B_{\epsilon}$(x)= { y$\in$ X | d(x,y) < $\epsilon$} 
It's not hard to show that the set of open balls of positive radius form a topology on a metric space. (Actually,to be more precise, they form a basis for a topology on X where each open set in the topology is a union of open balls. But let's not worry about that here.) 
Now let's recheck your examples in question 1. In (X,$T_{\epsilon}$), the open balls are the sets $B_{\epsilon}$(3)= { y$\in$ X | d(3,y) < $\epsilon$}. So yes,that's correct. 
Your second example is correct,but it needs clarification. Even though we have a Euclidean metric on the space, the space by definition only contains the integers in [1,4].Which means allowed balls in this space must have integer values of $\epsilon$. For example, $B_{\frac{1}{2}}$(3) = {3} since we're only allowed integers in the space. So only $\epsilon$ =1,2,3,4 are allowed here. So your answers are correct. 
What about (y,$T_d$)? A neighborhood is any subset of X that contains an open set that contains the point.{1,2,3} is certainly a nbh of 3, but is it the only one? Think carefully-doesn't [1,4]$\cap \mathbb Z$ = {1,2,3,4} also qualify under the definition? Sure,since {1,2,3}$\in T_d$ and {1,2,3} $\subset$ {1,2,3,4} =  [1,4]$\cap \mathbb Z$ ! Be careful! 
The same problem occurs with (X,$T_c$). Certainly, [1,4) is a nbh of 3,but what about X = [1,4]? Clearly, [1,4) $\subset $ [1,4]=X, so this too is a nbh of 3. 
That should be good enough for you to finish the rest yourself. I'll try and expand the problem tomorrow if you're still lost. Leave me a comment if you need more help! 
