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?