Prove that this definition of closed set yields a topology for $ S = [0, 1] $ Let $ S = [0, 1] $ and define a subset $ A $ of $ S $ to be closed if $ A = S $ or $ A $ is finite. Prove the following:
a) This definition of closed set yields a topology for $ S $.
Here is my approach: Let $ \mathcal{T} $ be the subset of $ S $ induced by the definition of closedness. Since $ \emptyset $ is finite and $ S = S $, we have $ \emptyset, S \in \mathcal{T} $. Next, suppose $ O_{1}, O_{2}, \dots \in \mathcal{T} $. If they are all finite, then their intersection is finite and hence in $ \mathcal{T} $. If one of them is $ S $ and the rest of them are finite, then their intersection is still finite. Either case, their intersection is always in $ \mathcal{T} $. Similarly, if they are all finite, then their union is finite and hence in $ \mathcal{T} $. If one of them is $ S $ and the rest of them are finite, then their union is $ S \in \mathcal{T} $. Either case, their union is always in $ \mathcal{T} $. Hence, $ \mathcal{T} $ is a topology.
b) $ S $ with this topology is connected, pathwise-connected, and compact, but $ S $ is not Hausdorff.
I have successfully proved the connected and pathwise-connected portion. To prove connectedness, first observe that clearly $ \emptyset $ and $ S $ are both open and closed. Suppose $ A $ is a nontrivial proper subset of $ S $. It suffices to prove that if $ A $ is closed, then $ A $ is not open. Since $ A $ is closed, $ A \in \mathcal{T} $. Thus, $ A $ is finite since $ A \neq S $. Hence, $ C(A) $ is not finite and $ C(A) \neq S $ since $ A $ is nonempty. This means that $ C(A) \notin \mathcal{T} $, so $ C(A) $ is not closed, so $ A $ is not open. The proof for connectedness completes. To prove path-connectedness, define $ f : [0, 1] \to S $ with $ f(x) = x $. We know that the identity map is continuous and $ f(0) = 0 $ and $ f(1) = 1 $. Therefore, $ S $ is pathwise-connected.
Can someone please check if my argument is logical and offer hint for the compact and Hausdorff part?
c) Each subset $ A $ of $ S $ is compact and that therefore, there are compact subsets of $ S $ that are not closed.
I am still struggling with this part. I know that a finite subset of a topological space is always compact. So if $ A $ is finite or $ A = S $, then $ A $ is compact, but I'm stuck on the case where $ A $ is not finite. 
 A: To prove path connectedness, you must prove that for all $x, y \in S$, there is a continuous function from $[0,1] \to S$ satisfying $f(0)=x$ and $f(1)=y$. This does not seem to be reflected in your argument.
Furthermore, you have not proved that the identity map from $[0,1] \to S$ is actually continuous. To do this, we have to show that for $A \subset S$ closed, $I^{-1}(A)=A$ is closed in the standard topology of $[0,1]$, but then if $A$ is closed in $S$, either $A$ is finite, so it is closed in the standard topology, or $A = S$, in which case it is also closed in the standard topology. So the identity map is continuous, and hence we can conclude that $S$ is path connected.
To see that $A \subset S$ is compact, let $\{U_i\}$ be an open cover of $A$. Then, the complement of each $\{U_i\}$ is finite, or $U_i =\emptyset$, in which case we throw it away.
Pick any $U_I$ , and note that $A = (U_I \cap A) \cup (C(U_I) \cap A)$. Since $C(U_I)$ is finite, $C(U_I) \cap A$ is finite, so $U_I \cap A$ (and hence $U_i$) will contain all but finitely many points of $A$, namely $a_1,a_2, ... a_n$, and since $\{ U_i\}$ is an open cover, we can find $U_1 ... U_n$ in the collection that contain $a_1 , ..., a_n$, so that the collection $\{U_I,U_1,...,U_n \}$ will cover $A$. Hence $A$ is compact. (And since $S \subset S$, $S$ is also compact).
Therefore, there exist compact subsets of $S$ that are not closed. 
$S$ is not Hausdorff. To see this, let $a,b \in S$, and $a \in A$, $b \in B$, where $A \cap B = \phi$ and $A,B$ are open in $S$. Then, note that $C(A) \cup C(B) = C(A \cap B) = S$ by De Morgan's Law, and since the union of two finite sets can't be infinite, either $A =S$ or $B=S$, but then their intersection can't be empty. So this gives a contradiction, and hence $S$ is not Hausdorff. 
