Prob. 2, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: The dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is metrizable.

Here's Prob. 2, Sec. 20 in the book Topology by James R. Munkres, 2nd edition:

Show that $\mathbb{R}\times \mathbb{R}$ in the dictionary order topology is metrizable.

The dictionary order on the set $\mathbb{R} \times \mathbb{R}$ is defined as follows:

For any two points $x_1\times y_1$, $x_2 \times y_2$ $\in \mathbb{R} \times \mathbb{R}$, we define $$x_1 \times y_1 \prec x_2 \times y_2$$ if and only if either $x_1 < x_2$, or if $x_1 = x_2$ and $y_1 < y_2$.

Now the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is the one having as a basis all sets of the fomm $$\left( a \times b, a \times c \right) \colon= \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ a \times b \prec x \times y \prec a \times c \ \right\},$$ where $a, b, c \in \mathbb{R}$ such that $b < c$.

Now if we define the funtion $d \colon \left(\mathbb{R} \times \mathbb{R} \right) \times \left( \mathbb{R} \times \mathbb{R} \right) \to \mathbb{R}$ as $$d\left( x_1 \times y_1, x_2 \times y_2 \right) \colon = \begin{cases} 1 \ \mbox{ if } \ x_1 \neq x_2; \\ \min \left(\ \vert y_1 - y_2 \vert, \ 1 \ \right) \ \mbox{ otherwise }, \end{cases} $$ then is this function $d$ a metric? How to verify the triangle inequality?

Does this function give the dictionary order topology on $\mathbb{R} \times \mathbb{R}$?

If $x_1 = x_2 = x_3$, then we have $$ \begin{align} & \ d\left(x_1 \times y_1, x_3 \times y_3 \right) \\ &= \min\left( \vert y_1 - y_3 \vert, \ 1 \right) \\ &\leq \min\left( \vert y_1 - y_2 \vert, \ 1 \right) + \min\left( \vert y_2 - y_3 \vert, \ 1 \right) \\ & \ \ \mbox{ [using the fact that this minimum is the same as the standard bounded metric on $\mathbb{R}$]} \\ &= d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align} $$ If $x_1 \neq x_2$ and $x_2 \neq x_3$, then we have $$ \begin{align} d\left(x_1 \times y_1, x_3 \times y_3 \right) &\leq 1 < 1 + 1 = d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align} $$ If $x_1 = x_2$ and $x_2 \neq x_3$, then $x_1 \neq x_3$ either, and so $$ \begin{align} d\left(x_1 \times y_1, x_3 \times y_3 \right) &= 1 \\ &\leq d\left(x_1 \times y_1, x_2 \times y_2 \right) + 1 \\ &= d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align}$$ And, similarly for the case when $x_1 \neq x_2$ and $x_2 = x_3$.

Is this demonstration of the triangle inequality complete and correct?

PS:

Assuming that the above $d$ is a metric, here is my attempt at showing that the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is indeed the one induced by the metric $d$ above.

Let $$B \colon= \{ a \} \times (b, c) = \{ \ a \times t \in \mathbb{R} \times \mathbb{R} \ \colon \ b < t < c \ \} = ( a \times b, a \times c) $$ be a basis element for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$, and let $x \times y \in B$. Then of course $x = a$ and $b < y < c$. Let us put $$ \epsilon \colon= \min \{ \ y-b, c-y, 1 \}. $$ Then if $s \times t \in B_d ( x \times y, \epsilon)$, then $s \times t \in \mathbb{R} \times \mathbb{R}$, and $$ d( s \times t, x \times y ) < \epsilon. \tag{1}$$ and, as $\epsilon \leq 1$, so $$ d( s \times t, x \times y ) < 1, $$ which implies that $s = x$, that is, $s = a$, and also from (0) we can conclude that $$ d( s \times t, x \times y ) = \min \{ \ \lvert t-y \rvert, 1 \ \}. $$ Then (1) implies that $$ d( s \times t, x \times y ) = \min \{ \ \lvert t-y \rvert, 1 \ \} < \epsilon = \min \{\ y - b, c - y , 1 \ \}. $$ So $$ d( s \times t, x \times y ) = \lvert t-y \rvert < \min \{ \ y-a, b-y \} , $$ The last relation implies that $b < t < c$. So $s \times t \in B$.

Thus for any basis set $B$ for the dictionary order topology and for any element $x \times y \in B$, we have a basis element $B_d ( x \times y, \epsilon)$ for the $d$-metric topology such that $$ x \times y \in B_d ( x \times y, \epsilon ) \subset B. $$ So the $d$-metric topology is finer than the dictionary order topology on $\mathbb{R} \times \mathbb{R}$.

Now let us consider an open ball $B_d( a \times b, \epsilon )$, where $a \times b \in \mathbb{R} \times \mathbb{R}$ and $\epsilon > 0$ are arbitrary. Let $x \times y \in B_d ( a \times b, \epsilon )$. Then if we choose a real number $\delta$ such that $$ 0 < \delta < \min \{ \ \epsilon - d( a \times b, x \times y), \ 1 \ \}, $$ then we note that $\delta < 1$ and also that $$ B_d ( x \times y, \delta ) \subset B_d ( a \times b, \epsilon ). \tag{2} $$

Now let us put $$ B \colon= \{ \ x \ \} \times ( y-\delta, y + \delta) = \big( \ x \times (y-\delta), \ x \times (y+ \delta) \ \big). $$ Then this $B$ is a basis element for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ such that $x \times y \in B$.

Moreover, if $s \times t \in B$, then $s = x$ and $y-\delta < t < y+\delta$ and hence $\lvert t-y \rvert < \delta$. But $\delta < 1$. So $$ d( s \times t, x \times y ) = \min \{ \lvert t-y \rvert, 1 \} = \lvert t-y \rvert < \delta, $$ which implies that $ s \times t \in B_d( x \times y, \delta )$ and hence also that $s \times t \in B_d( a \times b, \epsilon)$ by virtue of (2) above. Therefore $B \subset B_d( a \times b, \epsilon)$.

Thus we have shown that for any basis element $B_d( a \times b, \epsilon)$ for the $d$-metric topology and for any element $x \times y \in B_d( a \times b, \epsilon)$, there is a basis element $B$ for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ such that $$ x \times y \in B \subset B_d( a \times b, \epsilon). $$ Thus the dictionary order topology is finer than the $d$-metric topology.

The preceding few paragraphs show that the $d$- metric topology is the same as the dictionary order topology on $\mathbb{R} \times \mathbb{R}$.

Is this proof correct? Is each and every step of it correct in its logic and presentation? If not, then where lies the problem?

Here's an outline:

It might be enlightening to convince yourself that the dictionary order topology on $\mathbb{R}\times\mathbb{R}$ is the homeomorphic to the disjoint union of continuum many copies of $\mathbb{R}$.

One may check, and my recollection is that Munkres does this, that if $d$ is a metric, then $d' = \min(d,1)$ is a metric inducing the same topology. Thus, as far as the topology of metric spaces is concerned, it is entirely sufficient to consider metrics which are $\leq 1$.

Now, suppose that $(X_\alpha)_{\alpha \in I}$ are metric spaces and that the corresponding metrics $d_\alpha$ all have $d_\alpha \leq 1$. Form the disjoint union $X = \bigsqcup_{\alpha \in I} X_\alpha$ and give it its natural topology (open sets in $X$ are disjoint unions $\bigsqcup_{\alpha \in I} U_\alpha$ where $U_\alpha$ is open in $X_\alpha$). You may find it less distracting to check in this setting that $d$ defined by $d(x,y) = \begin{cases} d_\alpha(x,y) & \text{ if } x,y \in X_\alpha \\ 1 & \text{ otherwise} \end{cases}$ is a metric on $X$ and induces the aforementioned topology.

To put it in a slogan: "the disjoint union of metrizeable spaces is metrizeable".