Properties of the co-countable topology on $[0,1]$ I am trying to learn topology by myself. I have the following questions and my attempts of their proofs. Since I have no one to consult with, I am posting here. If anyone check my proofs that would be great. I know it is a bit long, my apologies for that. Many thanks for your time in advance!

Let $\tau$ be the system of sets consisting of $\emptyset$ and every subset of the the closed unit interval $[0,1]$ obtained by deleting a finite or countable number of points from $X$. Prove that
  
  
*
  
*$T=(X, \tau)$ is topological space 
  
*$T$ satisfies neither the second nor the first axiom of countability
  
*Only convergent sequences in $T$ are stationary sequences i.e., the sequences all of whose terms are the same starting from some index $n$
  
*$M=(0,1]$ has the contact point $0$ as a contact point, but contains no sequence of points converging to $0$
  

So, for (1):
$\emptyset\in\tau$ and if we delete $0$ points (finite) from $[0,1]$ then $X\in\tau$.
Let $\cup_\alpha G_\alpha$ be open sets, then $G_\alpha=X-F_\alpha$ where $F_\alpha$ is at most countable. Then $\cup_\alpha G_\alpha=\cup (X-F_\alpha)=X-\cap_\alpha F_\alpha$, but $\cap\alpha F_\alpha \subset F_\alpha$ is countable (subset of a countable set is countable). Thus arbitrary union of open sets is in topology. I can similarly show "finite intersection property".
For (2): My idea is if I can show this topology does not satisfy first axiom of countability, then it also fails to satisfy second one (this idea might be wrong, though). Then, if $x\in T$, there is no countable neighbourhood base at $x$ since we have deleted countable points. The problem is I am not sure whether this explanation is true or not, also it does not seem rigorous enough to me.
For (3): I don't understand this claim. For any stationary sequence, I can delete them from my system so there would be no stationary sequence in this topology, but then I can create uncountably many stationary sequence (choose any $x\in T$ as members of the sequence). So a bit confusing, but it is clear that a stationary sequence is convergent. That is, any neighbourhood $G$ of $x$ contains every point after certain index. For contradiction, take any non-stationary sequence $\left\{x_n\right\}\to x$. Take neighbourhood $G$ of $x$ such that after certain index, say $N$, every member of $\left\{x_n\right\}$ deleted, except for $x_n=x$. Than $G$ does not contain any member of $\left\{x_n\right\}$, except $x$.
For (4): Suppose $G$ is neighbourhood of $0$ such that $G\cap M=\emptyset$. Then $G=\left\{0\right\}$ which cannot be an open set in this topology $T$. By using ii), if $x_n\to 0$ then, it should be stationary i.e., $x_n=0$ for $n>N$, which is impossible.
 A: The outline for (1) appears fine.
For (2) you must show for any $x \in X$ and any given family $\{ U_n : n \in \mathbb{N} \}$ of open neighbourhoods of $x$ you can find an open neighbourhood $V$ of $x$ such that $U_n \not\subseteq V$ for all $n \in \mathbb{N}$.  As a hint, note that if $U$ is nonempty and $x \in U$, then $U \not\subseteq X \setminus \{ x \}$.  (You are correct that if $X$ is not first-countable, then it cannot be second-countable.)
The point of (3) is to show that if a sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ is not eventually constant, then it cannot converge to any point.  As a hint, suppose that $\langle x_n \rangle_{n \in \mathbb{N}}$ is not eventually constant.  Then the set $A = \{ x_n : n \in \mathbb{N} \}$ is countably infinite.  For each $x \notin A$ we can easily construct an open set $U$ which contains no elements of the sequence (so the sequence cannot converge to any point not belonging to $A$).  For each $x \in A$ we can make a small adjustment to $U$ to get an open neighbourhood $U^\prime$ of $x$ such that the sequence is not eventually always inside of $U^\prime$.
For (4), note that every open neighbourhood of $0$ must meet the set $M$, and so $0$ is a limit point (contact point) of $M$.  By the idea of proving (3) we can show that no sequence of points in $M$ converges to $0$.  (For any sequence of points in $M$, construct an open neighbourhood of $0$ containing no points of that sequence.)
