Create an explicit extension of the 2-adic metric to $\mathbb{R}$ in which at least $(1,\infty)$ is connected.

Create an explicit extension of the 2-adic metric to $\mathbb{R}$ in which at least $(1,\infty)$ is connected, $\lvert x\rvert\lvert y\rvert=\lvert xy\rvert$ and $|x+y|_2\leq\max\{|x|_2,|y|_2\}$.

Here is what I've tried so far:

This question discusses an extension of the 2-adic metric to the real numbers and states that one exists, and is furthermore used in Monsky's theorem - but doesn't define it.

Call this metric $\lvert\cdot\rvert_{\times}$

It seems to me the logical extension of the metric to e.g. $\mathbb{Q}_2[\sqrt{2}]$ is simply to allow $$\lvert\cdot\rvert_{\times}=2^{\tfrac{1}{2}}\text{ or possibly }2^{\pm\tfrac{1}{2}}$$ as well as integer powers of $2$.

I may be mistaken. And I'm unsure how to check whether this extension contradicts the required conditions. Perhaps by continuing in this direction all the real numbers can be connected?

Several things in your post seem to be based on misunderstandings.

1) What you call $2$-adic metric is usually and should be called the $2$-adic absolute value. This value $| \cdot |_2$ is then used to define what should really be called "$2$-adic metric", $d(x,y) = |x-y|_2$.

2) A value that satisfies your properties is nonarchimedean, the metric it defines is an ultrametric: $d(x,y) \le \max \{d(x,z), d(z,y)\}$. Every ultrametric space is totally disconnected, meaning that no subset of it which contains more than one point is connected. This is a nice little exercise in ultrametrics as well as something that is covered in any book about this subject.

This shows that what you are asking for is impossible. I would like to point out some further things that I am not sure you are aware of:

3) The $2$-adic absolute value is, first of all, defined on $\mathbb{Q}$. The completion of $\mathbb{Q}$ w.r.t. it is $\mathbb{Q}_2$. The $2$-adic absolute value naturally extends to $\mathbb{Q}_2$.

4) $\mathbb{Q}_2$ and $\mathbb{R}$ are very different fields. Neither of them is in any way contained in the other. One should think of them as going off in different directions from $\mathbb{Q}$.

5) To qualify the last point, some subsets of $\mathbb{Q}_2$ and $\mathbb{R}$ can be "matched", although already here one should be careful. As long as a number $\alpha$ is algebraic over $\mathbb{Q}$, there is a natural extension of the $2$-adic absolute value both from $\mathbb{Q}$ to $\mathbb{Q}(\alpha)$ and from $\mathbb{Q}_2$ to $\mathbb{Q}_2(\alpha)$ _{(EDIT: ...and up to $[\Bbb Q(\alpha):\Bbb Q]$ reasonable extensions of $|\cdot|_2$ to $\Bbb Q(\alpha)$, namely, one for each non-archimedean place of $\Bbb Q(\alpha)$ lying above $2$ (= one for each irreducible factor of the $\Bbb Q$-minimal polynomial of $\alpha$ over $\Bbb Q_2$; = one for each distinct prime factor in the prime factorisation of $(2)$ in the ring of integers of $\Bbb Q(\alpha)$. In the example $\alpha=\sqrt{2}$, there is only one such extension of the value, as the prime 2 totally ramifies.}

As in your example, of course the only reasonable extension of the absolute value to $\alpha = \sqrt c$ for $c \in \mathbb{Q}_2$ is $|\sqrt c| := \sqrt{|c|}$. Much more generally, an element $\alpha$ whose minimal polynomial over $\mathbb{Q}_2$ has degree $d$ is given the value $|N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha)|^{1/d}$. This is treated extensively in any book on $p$-adics, local fields etc.

6) In this way, one could explicitly extend the $2$-adic value from $\mathbb{Q}$ (not $\mathbb{Q}_2$, which is not contained in $\mathbb{R}$!) to the algebraic closure of $\mathbb{Q}$ in $\mathbb{R}$. But to extend it further (as in the theories you quote), one sort of arbitrarily extends it to transcendentals. There is absolutely no meaningful way to assign a $2$-adic value to the real numbers $\pi$ or $e$ or any other transcendental number. To be more clear, the axiom of choice guarantees that we can assign $2$-adic values to all of them, but none of our choices would be better than others. You can make them have whatever value you want, as long as they are compatible (in the sense that if $e^\pi$ would happen to be an algebraical expression in, say, $\pi, e$ and some other transcendental number, then that relation would restrict our choices). So to clarify: There is no explicit construction of such a value -- not because there is no such value, but because there are far too infinitely many, and none of them is more reasonable than all others. (Pick some transcendental number. You can choose it to have whatever absolute value $>0$ you like to begin with, but then when you extend it further, you would first have to make a vast number of compatible choices (to everything that is algebraic over that number), but even after that you still have continuum many transcendentals left. Again you pick one and assign some arbitrary value to it, then go on ...! And you can never explicitly list all these possibilities, actually, you have to make continuum many choices, so that inductive process would never even come close to doing it all.)

Although @TorstenSchoeneberg is right in the 6 points of his answer, I just wanted to point out that it is possible to extend any $p$-adic valuation to $\mathbb{R}$ even to $\mathbb{C}$. The construction is as follows: since there exists a field isomorphism $T:\mathbb{C}\to\mathbb{C}_p$ that extends the identity map in $\mathbb{Q}$, you can define the valuation $|\cdot|_*:\mathbb{C}\to\mathbb{R}$ by $|x|_*=|T(x)|_p$. It is not hard to see that $|x+y|_*\leq\max\{|x|_*,|y|_*\}$ for all $x,y\in\mathbb{C}$. Then, the restriction of $|\cdot|_*$ to $\mathbb{R}$ is a valuation on $\mathbb{R}$ that extends the $p$-adic valuation of $\mathbb{Q}$. Of course, this construction cannot be done explicitly since there is not an explicit formula for $T$. We only know about its existence. For a proof of the fact that $\mathbb{C}$ and $\mathbb{C}_p$ are isomorphic you can see page 83 of the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - 1978.