Can any norm be used in a product metric

Yes, you are correct and the answer is no.

Such a possible norm is $$ |(x,y)|:= \left\| \begin{pmatrix} x + y \\ 4(x-y) \end{pmatrix}\right\|_2. $$ Then we have $$|(0,1)|=\sqrt{17}> 2\cdot 2 = 2|(1,1)|.$$

The relevant criteria is not only that $|\cdot|$ is a norm on $\Bbb R^2$, but also that it is a norm that is monotone in each component. Then $D$ would be a metric on $X\times Y$.

This criteria is satisfied for all $p$-norms on $\Bbb R^2$.

Here's an abstract way to organize things. You can actually define metrics to take values in any ordered monoid, as follows:

An ordered monoid is a set $R$ equipped with both a partial order $\ge$ and a monoid operation $+ : R \times R \to R$ which is monotone with respect to product order, meaning that if $a \ge b$ and $c \ge d$ then $a + c \ge b + d$. Two simple examples are $\mathbb{R}_{\ge 0}$ equipped with $+$ and the usual ordering, and $\mathbb{R}_{\ge 0}$ equipped with $\text{max}$ and the usual ordering.

Definition: If $R$ is an ordered monoid, an $R$-metric space is a set $X$ equipped with a function $d : X \times X \to R$ such that $0 \ge d(x, x)$ (where $0$ is the identity in $R$) and satisfying the triangle inequality $d(a, b) + d(b, c) \ge d(a, c)$.

There are extra axioms you can impose if you want, like that $+$ is commutative, that $d$ is symmetric, or that if $d(x, y) = d(y, x) = 0$ then $x = y$ ("nondegeneracy"). This is the set of axioms that falls out of thinking about enriched categories, though.

Example. An $R$-metric space with $R = (\mathbb{R}_{\ge 0}, +, \ge)$ is a metric space in the usual sense but without the requirement that $d$ be symmetric or nondegenerate. These are sometimes called "quasipseudometric spaces."

Example. An $R$-metric space with $R = (\mathbb{R}_{\ge 0}, \text{max}, \ge)$ is an ultrametric space, but again without the requirement that $d$ be symmetric or nondegenerate. So I guess you could call them quasipseudoultrametric spaces but that would be truly terrible.

Example. A much weirder example but one that turns out to be relevant to thinking about the relationship between norms and metrics. If $R$ is taken to be a group equipped with the trivial ordering, then an $R$-metric space is exactly a torsor over $R$.

The point of introducing this formalism is to observe the following:

• If $X, Y$ are two metric spaces in the ordinary sense, $X \times Y$ inherits a canonical $R$-metric valued in $R = \mathbb{R}_{\ge 0}^2$ instead of $\mathbb{R}_{\ge 0}$, given by $(d_X, d_Y)$.
• If $f : R_1 \to R_2$ is a monotone lax monoid homomorphism between two ordered monoids (meaning $f(0) = 0, f(a) + f(b) \ge f(a + b)$, and if $a \ge b$ then $f(a) \ge f(b)$) then applying $f$ to an $R_1$-metric produces an $R_2$-metric ("base change").

The upshot is that we can, from these very simple abstract considerations, produce a metric on a product $X \times Y$ of two metric spaces from any map $f : \mathbb{R}_{\ge 0}^2 \to \mathbb{R}_{\ge 0}$ satisfying

• $f(0, 0) = 0$,
• $f(a_1, a_2) + f(b_1, b_2) \ge f(a_1 + b_1, a_2 + b_2)$, and
• if $a_1 \ge a_2$ and $b_1 \ge b_2$ then $f(a_1, a_2) \ge f(b_1, b_2)$.

(We need to require the nondegeneracy condition that if $f(a, b) = 0$ then $a = b = 0$ to send nondegenerate metrics to nondegenerate metrics, also.)

This formalism, even applied only to the special case of the ordered monoids $\mathbb{R}_{\ge 0}^n$, allows us to talk about convergence with respect to multiple metrics at once without needing to choose a way to combine them into a single metric. In fact they don't even need to be nondegenerate as long as they're jointly nondegenerate, and we can talk about infinitely many metrics at once, so for example the family of seminorms used to define a Frechet space can be thought of as inducing a single genuine metric, just taking values in $\mathbb{R}_{\ge 0}^{\infty}$ rather than $\mathbb{R}_{\ge 0}$.