I have to introduce a metric on $\mathbb{R}$ (edit/addendum: there are two $\mathbb{R}$ here, the domain and codomain of the function $f$ below, we are allowed to change the metric (from the standard metric) on either of them, or both.) that induces a standard topology, and such that $f:x \mapsto x^2$ becomes uniformly continuous.
We have our homeomorphism $\phi:(0,1) \rightarrow \mathbb{R}, \ x \mapsto \tan \pi \left(x - \frac{1}{2}\right)$. So the metric defined as follows $d(x,y) :=|\phi^{-1}(x) - \phi^{-1}(y)|$ gives us what we need. It induces a standard topology, since $\phi^{-1}$ is also a homeomorphism. And also makes $(\mathbb{R},d)$ a compact space since $(0,1)$ is compact and $\phi^{-1}$ is a homeomorphism.
If I'm able to show that $f$ acting on this constructed compact $\mathbb{R}$ is continuous, then by uniform continuit theorem, $f$ will be uniformly continuous. But I'm having trouble proceeding with that. Attempting a sequential continuity proof, I have to show that $|\arctan x_n - \arctan x| \rightarrow 0 \implies |x_n^2 - x^2| \rightarrow 0$
I don't know how to proceed, unable to see any manipulation that will lead me to my desired result. Can I get some kind of help?
Edit + Addendum: I seem to have made blundering mistakes in the way I posed the question, I sincerely apologise for that.
DanielWainfleet in his comments on the question has gracefully corrected my initial approach, which had faulty assumptions.
And there's also the answer posed by Connor Harris which works, since in the question we have '2 $\mathbb{R}$s', one in domain and codomain of $f: x \mapsto x^2$, and the metric introduced needn't be the same on both for ensuring the uniform continuity of $f$ as mentioned in the question. For the purposes of the question - $(\mathbb{R},d)$ as domain and $(\mathbb{R},d_{e})$ as codomain (using the notation used by Daniel Schepler in his comment on Connor Harris's answer) work. While, of course, independent of the required answer for the question I posed, Daniel Schepler's comment stands.
Define $$\phi(x) = \begin{cases} x^2 & x \geq 0 \\ -x^2 & x < 0 \end{cases}$$ and define $d(x, y) = |\phi(x) - \phi(y)|$. Then $|x^2 - y^2| \leq d(x, y)$, with equality iff $x$ and $y$ do not have opposite sign. This gives uniform continuity (in fact, Lipschitz continuity with constant 1).
EDIT: (This answer is wrong; see my discussion with Daniel Schepler in comments.)EDIT2: With the question as it's now been edited (to allow different metrics on $\mathbb{R}$ as domain and as codomain of $f$), this answer is fine.