Suppose $I^2 = [0, 1] × [0, 1]$ has the order topology generated by the dictionary order. Let $x$ be the point $(0.5, 0)$ in $I^2$ .
I want to show that there exists a sequence $(U_n)^\infty_{n=1}$ of open sets in $I^2$ so that $x ∈ U_n$ for every $n$, and that for every open set $V$ of $I^2$ containing the point $x$, we have $U_n\subset V$ for some $n$.
I know we can put a simple order relation on $\Bbb R^2$ as follows: $(a, b) < (c, d)$ if either (1) $a < c$, or (2) $a = c$ and $b < d$. And this is often called the dictionary order but I am not sure how to prove the existence of such sequence of open sets hmm~
You want a countable local base at a specific point of $I^2=[0,1] \times [0,1]$ in the topology induced from $<$, the lexicographic order on this square.
If we consider a point $(p,0)$ in $I^2$, where $p>0$ (so it's not the minimum, which has a slightly different neighbourhood base) then any neighbourhood of $(p,0)$ is contains some open interval $((a,b),(c,d))$ with $(a,b) < (p,0)$ and $(p,0) < (c,d)$. Between $(a,b)$ and $(p,0)$ (we must have $a < p$) we find some natural $m$ with $a< p-\frac1m < p$ and then $(p-\frac1m,0)$ lies inbetween $(a,b)$ and $(p,0)$. Also, we can always find some $n \in \mathbb{N}$ with $(p, \frac1n) < (c,d)$ (either $c=p$ and $d>0$ and take $m$ large enough, or always take $m=1$ if $c>p$).
It follows that the following set of open intervals is countable and as required:
$$\{((p-\frac1m, 0),(p, \frac1n)): n,m \in \mathbb{N}\}$$
Now take $p=\frac12$ as a special case. If you like you can specialise too to all open intervals using one variable: $((p-\frac1n,0),(p,\frac1n)), n \in \mathbb{N}$, taking the max of $n$ and $m$ in the previous case.
In fact around every point of $I^2$ we can do similar things and every point has a countable local base.