Prob. 4, Sec. 21, in Munkres' TOPOLOGY, 2nd ed: First countability of $\mathbb{R}_l$ and of the ordered square

Here is Prob. 4, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:

Show that $\mathbb{R}_l$ and the ordered square satisfy the first countability axiom. (This result does not, of course, imply that they are metrizable.)

My Attempt:

$\mathbb{R}_l$:

Let $p$ be any point of $\mathbb{R}_l$, the set of real numbers with the lower limit topology, which is the topology having as a basis all the closed-open intervals $[a, b) = \{ \ x \in \mathbb{R} \ \colon \ a \leq x < b \ \}$, where $a$ and $b$ are any real numbers such that $a < b$. For this point $p$, let us consider the countable collection $$\left\{ \ \left[p, p + \frac{1}{n} \right) \ \colon \ n \in \mathbb{N} \ \right\} \tag{1}$$ of basis neighborhoods of $p$ (i.e. basis open sets containing $p$).

We note that if $U$ is any open set in $\mathbb{R}_l$ and if $p \in U$, then there is some basis interval $[a, b)$ such that $$p \in [a, b) \subset U. \tag{2}$$

Now as $a \leq p < b$, so $0 < b-p$ and thus $\frac{1}{b-p} > 0$ also and we can find a natural number $N$ such that $N > \frac{1}{b-p}$. Then $\frac{1}{N} < b-p$ so that $p + \frac{1}{N} < b$, and thus the interval $$\left[ p, p+ \frac{1}{N} \right) \subset [ a, b),$$ and therefore $$p \in \left[ p, p+ \frac{1}{N} \right) \subset U,$$ by (2) above. Thus the collection in (1) is a countable basis at the point $p$.

But as $p$ was an arbitrary point of $\mathbb{R}_l$, so $\mathbb{R}_l$ has a countable basis at each of its points. Hence $\mathbb{R}_l$ satisfies the first countability axiom.

Is this proof correct? If so, then (how) can we conclude from this that $\mathbb{R}_l$ is not metrizable?

The Ordered Square:

Let $I \colon= [0, 1] = \{ \ x \in \mathbb{R} \ \colon \ 0 \leq x \leq 1 \ \}$. We note that the ordered square $I_0^2$ is the set $I \times I$ in the dictionary order topology.

Thus, given points $a \times b$ and $c \times d$ in $I \times I$, we define $a \times b \prec c \times d$ to mean that either $a < c$, or $a = c$ and $b < d$.

And, the dictionary order topology on $I \times I$ is the topology having as a basis all sets of each of the following three forms:

(i) $$[ \ 0 \times 0, a \times b \ ) \colon= \{ \ x \times y \in I \times I \ \colon 0 \times 0 \preceq x \times y \prec a \times b \ \},$$ where $a \times b \in I \times I$ such that $0 \times 0 \prec a \times b$;

(ii) $$( \ a \times b, 1 \times 1 \ ] \colon= \{ \ x \times y \in I \times I \ \colon a \times b \prec x \times y \preceq 1 \times 1 \ \},$$ where $a \times b \in I \times I$ such that $a \times b \prec 1 \times 1$;

(iii) $$( \ a \times b, c \times d \ ) \colon= \{ \ x \times y \in I \times I \ \colon \ a \times b \prec x \times y \prec c \times d \ \},$$ where $a \times b, c \times d \in I \times I$ such that $a \times b \prec c \times d$.

[How to write the MathJex code for formatting this as a list? I would appreciate if somebody around here can do this for me. ]

Now let $p \times q$ be an arbitrary point of $I_0^2$. We need to find a countable collection of basis sets containing $p \times q$ such that each open set containing $p \times q$ must contain at least one of these basis sets. There are several possible cases.

If $p \times q = 0 \times 0$, then a countable basis at $p \times q$ is the countable collection $$\left\{ \ \left[\ 0 \times 0 \ , 0 \times \frac{1}{n} \ \right) \ \colon \ n \in \mathbb{N} \ \right\}$$ of basis neighborhoods of $0 \times 0$.

If $0 < p \leq 1$ and $q = 0$, then $\frac{1}{p} > 0$ also and we can find a natural number $N$ such that $N > \frac{1}{p} > 0$, and so $0 < \frac{1}{N} < p$. Then the countable collection $$\left\{ \ \left( \ \left( p - \frac{1}{n} \right) \times 0 \ , \ p \times \frac{1}{n} \ \right) \ \colon \ n \in \mathbb{N}, \ n \geq N \ \right\}$$ is a required countable basis at $p \times q$.

If $0 < q < 1$, then $1-q > 0$ and so $\frac{1}{1-q} > 0$ also and we can find a natural number $N$ such that $$N > \max \left\{ \ \frac{1}{q}, \frac{1}{1-q} \ \right\}.$$ Then $N > \frac{1}{q} > 0$ and $N > \frac{1}{1-q} > 0$ and so $0 < \frac{1}{N} < q$ and $0 < \frac{1}{N} < 1-q$ and the latter inequality implies that $q < q + \frac{1}{N} < 1$. Thus the countable collection $$\left\{ \ \left( \ p \times \left( q - \frac{1}{n} \right) \ , \ p \times \left( q + \frac{1}{n} \right) \ \right) \ \colon \ n \in \mathbb{N}, \ n \geq N \ \right\}$$ of basis sets meets our requirement.

If $0 \leq p < 1$ and $q = 1$, then $1-p > 0$ and hence $\frac{1}{1-p} > 0$ also and we can choose a natural number $N$ such that $N > \frac{1}{1-p} > 0$; then $0 < \frac{1}{N} < 1-p$ and so $p < p+ \frac{1}{N} < 1$; therefore the countable collection $$\left\{ \ \left( \ p \times \left( 1 - \frac{1}{n} \right) \ , \ \left( p +\frac{1}{n} \right) \times 0 \ \right) \ \colon \ n \in \mathbb{N}, \ n \geq N \ \right\}$$ is a required countable basis at $p \times q$.

If $p \times q = 1 \times 1$, then a countable basis at $p \times q$ is the collection $$\left\{ \ \left( \ 1 \times \frac{1}{n+1} \ , \ 1 \times 1 \ \right] \ \colon \ n \in \mathbb{N} \ \right\}.$$

Is this demonstration correct? Have I managed to exhaust all possible cases? And if so, then have I managed to come up with a correct countable basis in each case?

How to show that $I_0^2$ is not metrizable?

The ideas for the first countability are fine. The proofs are fine too (although as usual in your questions, very detailed, at this level of courses it's OK to say that for any $x>0$ we have some $n$ with $0 < \frac{1}{n} < x$ without a detailed proof of that fact every time you use it, to give an example.)
The fact that both spaces $\mathbb{R}_l$ and $I_0^2$ are first countable doesn't show they are non-metrisable, on the contrary it shows how close these spaces are to metrisable spaces, without being metrisable:
$\mathbb{R}_l$ is first countable, (hereditarily) separable, (hereditarily) Lindelöf but not metrisable as it is not second-countable (in a metrisable space being separable implies being second countable, here it does not, alternatively note that its square is not normal as a proof of non-metrisability; the square of a metrisable space is metrisable and thus normal).
$I_0^2$ is likewise first countable, compact, but not separable (in a metrisable space compact implies separable, here it does not), as all open sets $((x,0),(x,1))$ are disjoint and non-empty and must all contain a point from any dense set, so all dense sets have size at least $\mathfrak{c}$)