Show that each component of $X = \{a, b, y_1, y_2, \ldots\}$ is Hausdorff, but $X$ is not Hausdorff. Given $X = \{a, b, y_1, y_2, \ldots\}$, where $a$, $b$, $y_i$ are points.
The topology on $X$ is $\tau = \{X\} \cup T_f\cup T_d$, where $T_f$ is the cofinite topology on $X$ and $T_d$ is the discrete topology on $Y = \{y_1, y_2, \ldots\}$.
How do I show that each component of $X = \{a, b, y_1, y_2,\ldots\}$ is Hausdorff, but $X$ is not Hausdorff?

I'm pretty sure I'm not doing this right, but if this were an exam question and I'm desperate to write something down:
To show that $X$ is not Hausdorff, I need to show that there exists two nonempty sets in $T$ that intersect. Since $T_d\subseteq T$ and $T_d$ contains all subsets of $Y$, then obviously we can find two subsets of $Y$ in $T$ that intersect.
To show that every component of $X$ is Hausdorff, since $a$, $b$, $y_1$, etc are points, the singleton sets of $X$ are also the components, which are Hausdorff.
It feels kind of stupid :( Help please?
 A: It is not true that if there are two nonempty open sets of $X$ that intersect then $X$ is not Hausdorff. In fact, any topological space that has a topology different from the indiscrete topology satisfies this condition, since there is always an open set $\mathscr{O}$ different from $X$ and $\varnothing$, and then $X\cap \mathscr{O}=\mathscr{O}\neq\varnothing$. So you would be saying that the only Hausdorff space is the space with one element (with its unique topology). That's certainly not true, so you've got the condition wrong.
It is true that if every pair of nonempty open sets in $X$ intersect, then $X$ is not Hausdorff. But the condition, while sufficient to establish non-Hausdorfness, is not necessary. And in your space, that condition does not hold, since $\{y_1\}$ and $\{y_2\}$ are both elements of $T_d$, and hence open, and they are disjoint.
Instead, to show that it is not Hausdorff we need to show that there exist $r,s\in X$, $r\neq s$, such that if $\mathscr{U}$ is any open set containing $r$ and $\mathscr{V}$ is any open set containing $s$, then $\mathscr{U}\cap\mathscr{V}\neq\varnothing$. This is just the negation of the definition of being Hausdorff ("for every $r,s\in X$, $r\neq s$, there exist open set $\mathscr{U}$ and $\mathscr{V}$ such that $r\in\mathscr{U}$, $y\in\mathscr{V}$, and $\mathscr{U}\cap\mathscr{V}=\varnothing$").
What are the open sets that contain $a$? The only open sets that contain $A$ are $X$ and elements of $T_f$ that contain $a$; that is, cofinite subsets of $X$ that contain $a$. What are the open sets that contain $b$? Again, the only ones are the cofinite subsets of $X$ that contain $b$.
So, if $X$ is infinite (presumably, you never actually say so!), then any two cofinite subsets intersect, so any open sets, one that contains $a$ and the other that contains $b$, intersect. So there are no disjoint neighborhoods of $a$ and $b$, hence $X$ is not Hausdorff if it is infinite.
(Conversely, if $X$ is finite, then the topology described is just the discrete topology, since the cofinite topology on a finite set is discrete; but the discrete topology is Hausdorff, so $X$ would be Hausdorff in that case. That is: $X$ is not Hausdorff if and only if it is finite.)
Now, to show that every connected component of $X$ is Hausdorff, note that each $\{y_i\}$ is a connected component. Indeed, $\{y_i\}\in T_d$, so it is open, and $X-\{y_i\}\in T_f$, so $\{y_i\}$ is also closed. Since it is clopen and a singleton, it is a component. Of course, being a singleton, it is Hausdorff in the induced topology (by vacuity).
Finally, $\{a,b\}$ is not a component, since the induced topology is discrete: there are open sets $\mathscr{U}$ and $\mathscr{V}$ of $X$ such that $\mathscr{U}\cap\{a,b\} = \{a\}$ and $\mathscr{V}\cap \{a,b\}=\{b\}$; (e.g., $\mathscr{U}=X-\{b\}$, $\mathscr{V}=X-{a}$). So $\{a\}$ and $\{b\}$ are the final two components of $X$, and again they are Hausdorff because they are singletons. 
