Quotients of the Ordered Square Let $[0,1]^2$ be the ordered square; i.e. it has the order topology given by the dictionary order.  This is a first countable compact space.  Let $\Delta=\{(x,x)\mid x\in [0,1]\}$.  Then is the quotient space $[0,1]^2/\Delta$ first countable?  
It isn't hard to see that there is a countable basis at every point of $[0,1]^2/\Delta$ that isn't the equivalence class of the diagonal $\Delta$, but I can't seem to figure out whether the equivalence class $[\Delta]$ has a countable basis or not.  I feel inclined to say that it does not, as by the fact that $[0,1]^2/\Delta$ is a quotient space, if $q:[0,1]^2\rightarrow [0,1]^2/\Delta$ is the quotient map that sends each element to its equivalence class, $U\subset [0,1]^2/\Delta$ is open iff $p^{-1}(U)$ is open.  In particular, $U$ is a neighborhood of $[\Delta]$ in $[0,1]^2/\Delta$ iff $p^{-1}(U)$ is an open set containing $\Delta$.  
EDIT: $\Delta$ is not closed.
If this quotient space is first-countable, do there exist quotient spaces of $[0,1]^2$ that are not first-countable?

Let $q:[0,1]^2\to[0,1]^2/\Delta$ be the quotient map, and let $p$ be the point in the quotient that corresponds to $\Delta$. Suppose that $U\subseteq[0,1]^2/\Delta$ is an open nbhd of $p$; then $q^{-1}[U]$ must be an open nbhd of $\Delta$ in the lexicographically ordered square. This means that there must be functions $f_U:(0,1)\to(0,1)$ and $g_U:(0,1)\to(0,1)$ such that:


*

*$f_U(x)<x<g_U(x)$ for each $x\in(0,1)$, and  

*$\{x\}\times(f_U(x),g_U(x))\subseteq q^{-1}[U]$ for each $x\in(0,1)$.


Use these functions $g_U$ and $f_U$ to show that if $\{U_n:n\in\Bbb N\}$ is any countable family of open nbhds of $p$, there is an open nbhd $V$ of $p$ such that $U_n\setminus V\ne\varnothing$ for each $n\in\Bbb N$; this shows that no countable family of open nbhds of $p$ can be a base at $p$.
