Relation between metric and uniform convergence I have a question about a relation between metric and uniform convergence on $\mathbb{R}$

Question
It is true that the extended real $\overline{\mathbb{R}}$ is completely metrizable.
Let $\rho$ be a metric which completely metrizes $\overline{\mathbb{R}}$
Let $d$ be a metric on $\mathbb{R}$.
Let $E$ be a set and $f_n:E\rightarrow \mathbb{R}$ be a sequence of functions which converges uniformly to $f$ under $d$.
Then, does $f_n$ converge uniformly to $f$ under $\rho$?

If this is true, i want to verify whether the following generalization is true:

Let $X$ be a topological space which is metrizable by either $d$ or $\rho $.
Let $E$ be a set and $f_n:E\rightarrow X$ be a sequence of functions converges uniformly to $f$ under $\rho$.
Then, does $f_n$ converge uniformly to $f$ under $d$?

 A: I assume all metrics shall induce the standard topologies on $\mathbb{R}$ resp. $\overline{\mathbb{R}}$.
The answer is negative.
Consider the map
$$T\colon x\mapsto \frac{1+ix}{1-ix}$$
which maps $\mathbb{R}$ bijectively to the unit circle minus the point $-1$, and let $d(x,y) = \lvert T(x) - T(y)\rvert$. Since $T$ as well as its inverse are continuous ($T$ is a Möbius transformation), $d$ induces the standard topology on $\mathbb{R}$. But $d$ has the property that a ball around a large enough positive number contains all negative numbers of sufficiently large absolute value (and vice versa), which is not compatible with a metric on $\overline{\mathbb{R}}$. Let
$$f_n(x) = \begin{cases}x &, \lvert x\rvert \leqslant n \\ -x &, \lvert x\rvert > n. \end{cases}$$
Then $f_n$ converges uniformly to the identity with respect to $d$, since for $\lvert x\rvert > n$ we have
$$d(x,-x) = \left\lvert \frac{1+ix}{1-ix} - \frac{1-ix}{1+ix}\right\rvert = \frac{\lvert (1+ix)^2 - (1-ix)^2\rvert}{1+x^2} = \frac{4\lvert x\rvert}{1+x^2} < \frac{4}{n}.$$
But
$$\sup_{x\in\mathbb{R}} \rho(f_n(x),x) \geqslant \sup_{\lvert x\rvert > n} \rho(-x,x) \geqslant \lim_{\lvert x\rvert\to\infty} \rho(-x,x) = \rho(-\infty,+\infty),$$
so $(f_n)$ cannot converge uniformly with respect to $\rho$.
