Can someone please verify my proof or offer suggestions for improvement?
I am aware that there is a similar question elsewhere, but I want help with my proof in particular.
Let $X$ be a metric space with metric $\operatorname{d}$. Show that $\operatorname{d}:X \times X \longrightarrow \mathbb{R}$ is continuous.
Let $U \subseteq \mathbb{R}$ be open. Assume $\operatorname{d}^{-1}(U) \neq \varnothing$.
Let $(x, y) \in \operatorname{d}^{-1}(U)$ and set $\epsilon_1 = \frac{1}{2}(b-\operatorname{d}(x, y))$
Clearly, if $(\alpha, \beta) \in B_d(x, \epsilon_1) \times B_d(y, \epsilon_1)$, then \begin{eqnarray} \operatorname{d}(\alpha, \beta) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(x, y) + \operatorname{d}(y, \beta)\\ &<& \operatorname{d}(x, y) + 2\left(\frac{b-\operatorname{d}(x, y)}{2}\right) \\ \end{eqnarray}
That is,
$$\operatorname{d}(\alpha, \beta) < b$$
Similarly, set $\epsilon_2 = \frac{1}{2}(\operatorname{d}(x, y) - a)$.
Let $(\alpha, \beta) \in B_d(x, \epsilon_2) \times B_d(y, \epsilon_2)$. Assume, for the sake of contradiction, that $\operatorname{d}(\alpha, \beta) \leq a$.
Then,
\begin{eqnarray} \operatorname{d}(x, y) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(\alpha, \beta) + \operatorname{d}(\beta, y) \\ \operatorname{d}(x, y) &<& \operatorname{d}(x, y)-a + \operatorname{d}(\alpha, \beta) \\ &<& \operatorname{d}(x, y) \end{eqnarray}
Therefore, it must be the case that $$\operatorname{d}(\alpha, \beta) > a$$
Set $\epsilon = \operatorname{min}\{\epsilon_1, \epsilon_2\}$.
Then, if $(\alpha, \beta) \in B_d(x, \epsilon) \times B_d(y, \epsilon)$,
$$a < \operatorname{d}(\alpha, \beta) < b$$
Since $\operatorname{d}^{-1}(U)$ can be written as the union of open sets in the metric topology, it follows that $\operatorname{d}^{-1}(U)$ is open. Therefore, $d$ is continuous.