Convergence in the product topology iff mappings converge 
Let $x_1,x_2,\ldots$ be a sequence of points of the product space $\prod X_\alpha$.  Show that this sequence converges to a point $x$ if and only if the sequence $\pi_\alpha(x_1),\pi_\alpha(x_2),\ldots$ converges to $\pi_\alpha(x)$ for each $\alpha$.

$\Rightarrow$  Suppose that $x_n \to x$.  The $\pi_\alpha(x_n) \to \pi_\alpha(x)$ since each of the $\pi_\alpha$ are continuous.
$ \Leftarrow $ Suppose that $\pi_\alpha(x_n) \to \pi_\alpha(x)$ for each $\alpha$.  Let $ x \in \prod_\alpha U_\alpha$.  Then for some natural number $N$, for $n \geq N$ we have $x_n \in \prod_\alpha U_\alpha$.  Now we have $U_\alpha \neq X_\alpha$ for only finitely many $\alpha$, say $\alpha_1,\ldots,\alpha_k$.  For each $j \in 1,\ldots,k$ let $N_j$ be such that $\pi_{\alpha_k}(x_n) \in U_{\alpha_j}$ for all $n \geq N_j$.  Let $\overline{N}=\max\{N_1,\ldots,N_k\}$.  Then $x_n \in \prod_\alpha U_\alpha$ for $n \geq \overline{N}$.
I'm stuck on the $\Leftarrow$.  I tried to argue through...but I'm having problems making my argument eloquent (and I am not confident that it is correct).
 A: Your argument is worded just a little clumsily, but the basic idea is just fine. Here’s one way to clean it up a bit.
Let $A$ be the index set, and suppose that $\langle\pi_\alpha(x_n):n\in\Bbb N\rangle\to\pi_\alpha(x)$ for each $\alpha\in A$; we want to show that $\langle x_n:n\in\Bbb N\rangle\to x$. To this end let $U$ be any open nbhd of $x$ in $X=\prod_{\alpha\in A}X_\alpha$. Then there is a finite $F\subseteq A$ and open sets $V_\alpha\subseteq X_\alpha$ for $\alpha\in A$ such that $V_\alpha=X_\alpha$ for all $\alpha\in A\setminus F$, and $$x\in\prod_{\alpha\in A}V_\alpha\subseteq U\;.$$
For each $\alpha\in F$ there is an $m_\alpha\in\Bbb N$ such that $\pi_\alpha(x_n)\in V_\alpha$ whenever $n\ge m_\alpha$; $F$ is finite, so let $m=\max\{m_\alpha:\alpha\in F\}$. Then for any $n\ge m$ we have $\pi_\alpha(x_n)\in V_\alpha$ for all $\alpha\in A$ and hence $$x_n\in\prod_{\alpha\in A}V_\alpha\subseteq U\;.$$ $U$ was an arbitrary open nbhd of $x$, so $\langle x_n:n\in\Bbb N\rangle\to x$.
