a) Hausdorff implies $T_1$ which implies singletons are closed. Note that closure of $\{x\}$ is defined: $\bigcap\{F:x\in F\}$, and this must equal $\{x\}$ by the aforementioned property of $T_1$ spaces.
b) Choose a point $y \not= x$. We note that there is an open set containing $x$, that does not contain $y$ by choosing 2 open sets, one containing $x$, the other containing $y$, such that the sets are disjoint. The result follows - if $y$ is not in one open set containing $x$, it is certainly not in every open set containing $x$.
c) a: Use the cofinite topology on the set of reals. Singletons are again closed. The fact that every open set intersects every other open set (except $\emptyset$) shows that this topology is not Hausdorff.
b: Use the cofinite topology on the reals. Consider open sets of the form $\{r\}^c$ where $r$ is a real number different from $\{x\}$. The intersection of all these sets is {x} but again, the space is not Hausdorff.
Part 2: Edit: This construction doesn't work, I can't think of a way to fix it sorry. Let $H$ be a Hausdorff space. Let's construct the discrete space inductively.
Induction
Suppose we have the set of points $\{x_i : 1\le i \le n\}$ and a corresponding collection of open sets $O_i$ such that
1.$x_i \in O_i$
2. $x_j \not \in O_i$ where $j \not= i$
3. $\bigcup_{i=1} ^n O_i$ does not cover $H$.
Choose points $x_{n+1}$ and $y$ in $[\bigcup_{i=1}^n O_i]^c$. Choose an open set $O_{n+1}$ around $x_{n+1}$ that does not contain $y$ or $x_i : 1\le i \le n$. (We can find this set by intersecting $n+1$ open sets that contain $x_{n+1}$ which are disjoint from the $x_i : 1\le i \le n$, and $y$. Now we have set of points $\{x_i : 1\le i \le n+1\}$ and a corresponding collection of open sets $O_i$ such that
1. $x_i \in O_i$
2. $x_j \not \in O_i$ where $j \not= i$
3. $\bigcup_{i=1} ^{n+1} O_i$ does not cover $H$.
Base Case
Chose $x_1$ and $y$ in $H$, and choose $O_1$ which contains $x_1$ but not $y$.
Conclusion
Now we have shown the existence of an increasing sequence of subspaces $S_n = \{x_i: 0 \le i \le n\}$ which are discrete. Just consider that $\{x_i\} = S_n \cap O_i \Rightarrow \{x_i\} $ is open in the subspace topology. (By increasing sequence I mean $S_n \subset S_{n+1}$) It remains to be shown that the limit is discrete.
Let $S_{\infty} = \bigcup_n S_n = \{x_i : 1\le i\}$. Choose any point $x_i$ in $S_{\infty}$. Note that, by construction, $O_i \cap S_{\infty} = \{x_i\}$ so all singletons are open and therefore $S_{\infty}$ is a discrete subspace which is countable and infinite.