Let $(x_\alpha)_{\alpha\in J}$ be a net in a topological space $X$. Show that if $J$ is a finite set then the net converges? ...
Why does it have to converge?
I don't understand why with nets this is true, whilst with sequences it isn't. if the value alternates between two values shouldn't it never converge?
By the definition of the concept of "net", $J$ must be a directed set. Being finite, it must have a unique maximal element $j_\infty$, i.e. $j \le j_\infty \ \forall j \in J$ (intuitively, you can "reach" infinity, which you cannot do in $\Bbb N$). Then $x_{j_\infty}$ is the limit of the net.
EDIT:
To clarify the question in the comment: let $J = \{1, 2, \dots, 10\}$ with the usual order. Let $x_1 = x_3 = \dots = x_9 = 0$ and $x_2 = x_4 = \dots = x_{10} = 1$. Let us show that $0$ cannot be the limit of this net.
Take a neighbourhood $V$ of $0$, small enough such that $1 \notin V$. The definition of "limit of a net" says that there should exist $n_V \in J$ such that for every $n \ge n_V$ we should have $x_n \in V$. Notice that no matter who $n_V$ would be, clearly $10 \ge n_V$. But do we have $1 = x_{10} \in V$? Of course not, by how we chose $V$. This shows that $0$ does not satisfy the definition of "limit of a net".
On the other hand, $1$ does satisfy the definition: choose any neighbourhood $V$ of $1$ and always choose $n_V = 10$. Then all the terms with index greater than $10$ (which in fact means just $x_{10}$) are in $V$, and you just don't care about the first $9$ terms.