Step-by-step explanation:
If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are [tex]x_1, x_2, ..., x_n[/tex]. So there are disjoint open neighborhoods [tex]U_{12}[/tex] and [tex]U_2[/tex], of [tex]x_1[/tex] and [tex]x_2[/tex] respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods [tex]U_{13}[/tex] and [tex]U_3[/tex] of [tex]x_1[/tex] and [tex]x_3[/tex] respectively, and disjoint open neighborhoods [tex]U_{14}[/tex] and [tex]U_4[/tex] of [tex]x_1[/tex] and [tex]x_4[/tex], and so on, all the way to disjoint open neighborhoods [tex]U_{1n}[/tex], and [tex]U_n[/tex] of [tex]x_1[/tex] and [tex]x_n[/tex] respectively. So [tex]U=U_2 \cup U_3 \cup ... \cup U_n[/tex] has every element of [tex]X[/tex] in it, except for [tex]x_1[/tex]. Since [tex]U[/tex] is union of open sets, it is open, and so [tex]U^c[/tex], which is the singleton [tex]\{ x_1\}[/tex], is closed. Therefore every singleton is closed.
Now, remember finite union of closed sets is closed, so [tex]\{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\}[/tex] is closed, and so its complemented, which is [tex]\{ x_1\}[/tex] is open. Therefore every singleton is also open.
That means any two points of [tex]X[/tex] belong to different connected components (since we can express X as the union of the open sets [tex] \{ x_1\} \cup \{ x_2,...,x_n\}[/tex], so that [tex]x_1[/tex] is in a different connected component than [tex]x_2,...,x_n [/tex], and same could be done with any [tex]x_i[/tex]), and so each point is in its own connected component. And so the space is totally disconnected.
