Prove $\mathbb R_\ell$ and the ordered square are first-countable. $\mathbb R_\ell$ is the lower-limit topology, which is first countable.
Proof:
Let $x \in \mathbb R$. The collection $U = \{[x, x + \frac{1}{n})\}_{n \in \mathbb Z_+}$ of basis elements is a countable basis about $x$, since for any basis element $[a,b)$ containing $x$, by the Archimedian property there is some $k \in \mathbb Z_+$ such that $x + \frac{1}{k} \in [a,b)$. 
The ordered square $I^2_0$ is first countable.
Proof:
Denote an element of $I^2_0$ by $(x \times y)$. A basis element of $I^2_0$ is of the form $\{x\} \times (s,t)$. Let $(a \times b) \in I^2_0$. We do casework to find a countable basis.
If $0 < a < 1$, $0 < b < 1$ then any neighborhood $(a \times b)$ contains some $\{a\} \times (b-1/n, b/n)$ for some $n$.
If $(a \times b) = (0 \times 0)$, any neighborhood of $(a \times b)$ contains $\{0\} \times [0, 1/n)$ for some $n$.
If $(a \times b) = (0 \times 1)$, any neighborhood of $(a \times b)$ contains $\{0\} \times (1 - 1/n, 1]$ for some $n$.
If $(a \times b) = (1 \times 0)$, any neighborhood of $(a \times b)$ contains $\{1\} \times [0, 1/n)$.
Finally, if $(a \times b) = (1 \times 1)$, any neighborhood of $(a \times b)$ contains $\{1\} \times (1 - 1/n, 1]$.
The collection containing the sets of the form presented in the casework is a countable basis.
Are there any errors? I think some details are missing in the casework of the second proof, but I can't pinpoint them exactly. If there are no mistakes, any improvements? The proof I've presented is admittedly quite messy, but I didn't see a better way to approach the problem.
EDIT: First version of the first proof had $[x, (n+1)x/n)$ which didn't account for $x <0$. The current version should work.
 A: For $\Bbb R_l$ the Archimedean principle is invoked (which is correct) but you don't show how it is applied, so depending on the level of detail expected, you might want to fill in that detail.
You just need the simple lemma

If $ x \in \Bbb R, x >0$ there is some $n \in \Bbb N$ such that $\frac{1}{n} < x$.

which is an almost immediate consequence of the AP.
For the lexicographically ordered square $I^2$, there are essentially three cases: $x=(0,0)$ the minimal element, and this point has (by definition of the order topology) basic neighbourhoods of the form $[(0,0), (a,b))$ where $(a,b) > (0,0) \in I^2$. And any such neighbourhood contains one of the form $[(0,0), (0,\frac1n))$, $n \in \Bbb N$. Trivially if $a>0$ and by the above lemma if $a=0, b>0$.
The other special case if $x=(1,1)$, the maximal element, which has basic neighbourhoods of the form $((a,b), (1,1)]$ with $(a,b) < (1,1)$ in $I^2$. This contains a basic neighbourhood of the form $((1,1-\frac1n),(1,1)]$, again trivially if $a < 1$ and by the AP if $a=1, b < 1$. 
Other points $(x,y)$ that are not $(0,0)$ or $(1,1)$ can have three different forms, but all have basic neighbourhoods of the form $((a,b), (c,d))$ with $(a,b) < (x,y) < (c,d)$. 
subcase a: $1 \ge x>0$ and $y=0$. Here we can use $((x-\frac1n,1),(x,\frac1n))$ for some $n$. Check this.
subcase b: $0 \le x<1$ and $y=1$. Here we can use $((x,1-\frac1n),(x+\frac1n,0))$ for some $n$. Check this too.
subcase c:  $0 < y < 1$, here we can use $((x,y-\frac1n),(x,y+\frac1n))$ for some $n$. 
So all $(x,y)$ have a countable local base. Still some small details for you to verify..
