Here is Prob. 6, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that $\mathbb{R}_\ell$ and $I_0^2$ are not metrizable.
My Attempt:
First, we consider $\mathbb{R}_\ell$. Recall that $\mathbb{R}_\ell$ is the set of real numbers with the so-called "lower limit topology" having as a basis the collection $$ \big\{ \, [a, b) \, \colon \, a \in \mathbb{R}, b \in \mathbb{R}, a < b \, \big\} $$ of all the (bounded) closed-from-the-left-open-from-the-right intervals on the real line.
Every metrizable Lindelof space has a countable basis, by Prob. 5 (b), Sec. 30, in Munkres. Here and here is my Math SE post on this very problem.
However, $\mathbb{R}_\ell$ is a Lindelof space that has no countable basis. Please refer to Example 3, Sec. 30, in Munkres.
Therefore $\mathbb{R}_\ell$ is not metrizable.
Am I right?
Now we consider $I_0^2$. Recall that $I_0^2 = [0, 1] \times [0, 1]$ with the dictionary order topology.
The topological space $I_0^2$ is a simply ordered set that has the least upper bound property; so by Theorem 27.1 in Munkres every closed interval $[ a\times b, c \times d]$ in $I_0^2$ is compact, where $a \times b \in I_0^2$, $c \times d \in I_0^2$ and $$ a \times b <_{I_0^2} c \times d. $$ In particular, we have $$ I_0^2 = [0 \times 0, 1 \times 1 ] $$ so that $I_0^2$ itself is compact.
As has been shown in Example 5, Sec. 30, in Munkres, the subspace $A \colon= [0, 1] \times (0, 1)$ of $I_0^2$ is not Lindelof.
Since every second-countable space is Lindelof, by Theorem 30.3 (a) in Munkres, and since the subspace $A = [0, 1] \times (0, 1)$ of $I_0^2$ is not Lindelof, therefore we can conclude that $A$ is not second countable either.
But since every subspace of a second countable space is also second-countable, by Theorem 30.2 in Munkres, and since the subspace $A = [0, 1] \times (0, 1)$ of the topological space $I_0^2$ is not second countable, therefore we can conclude that $I_0^2$ itself is not second-countable either. Thus $I_0^2$ has no countable basis.
Thus $I_0^2$ is a compact space having no countable basis.
But every compact metrizable space has a countable basis, by Prob. 4, Sec. 30, in Munkres; here is my Math SE post on this very problem.
Finally since every compact metrizable space has a countable basis and since $I_0^2$ is a compact space that has no countable basis, therefore we can conclude that $I_0^2$ is not metrizable.
Am I right?
Is each one of the two parts of my proof correct as a whole? If so, is each and every detail correct as well? Is my presentation in both parts clear enough? If not, then where are the problem areas?
