Correlation of X with signum(Y) where X and Y are normal standard with a given correlation Setup:
$$
X\sim \mathcal{N}(0, 1)\\
Y\sim \mathcal{N}(0, 1)\\
Corr(X,Y)=ρ
$$
From simulations I see $ρ(Y, signum(X))$ is fairly close to $0.8ρ$, suspect is a proxy for ${\sqrt{(2/\pi)}}ρ$, am wondering if this is indeed a general case and if so looking for some analytical proof or clues to the proof.
 A: Here is a counterexample, which takes advantage of the fact you have not specified the relationship between $X$ and $Y$ beyond their correlation.
Let $X \sim N(0,1)$ and let $Y=X$ if $|X|<1.538172254455$ while $Y=-X$ if $|X|\ge 1.538172254455$ (the square root of the median of a $\chi^2_3$ distribution, but that is not important here).  Clearly by symmetry $Y\sim N(0,1)$ too.
It turns out $\text{Cor(X,Y)} \approx 0$ but $\text{Cor}(Y,\text{signum}(X)) \approx 0.309$
Here is some R code to confirm this:
set.seed(2021)
X <- rnorm(10^5)
Y <- ifelse(abs(X) < 1.538172254455, X, -X)
cor(X, Y)
# 0.002905442
cor(Y, sign(X))
# 0.3107578
2*dnorm(0)-4*dnorm(sqrt(qchisq(0.5,3))) # theoretical expectation
# 0.3090011


If $X=Y$ then here $\text{Cor}(X,\text{signum}(X))=E[|X|]$ the mean of a half-normal distribution which is indeed $\sqrt{\frac{2}{\pi}}$.
Both $X$ and $\text{signum}(X)$ have mean $0$ and variance $1$ so $\text{Cor}(X,\text{signum}(X)) $ $= \text{Cov}(X,\text{signum}(X)) $ $= E[X\text{signum}(X)] $ $=E[|X|]$.  And it is easy to show $E[|X|]=2\int_0^\infty x \frac1{\sqrt{2\pi}}\exp(-x^2/2)\, dx = \sqrt{\frac2\pi}$ - just differentiate $\exp(-x^2/2)$ to see what is going on.
I suspect you may have to assume something extra for your result to avoid things like my counter-example, for example assuming $X$ and $Y$ to be bivariate normal.  If they are, then there you can let  $Z=\frac{Y-\rho X}{\sqrt{1-\rho^2}}$, and you will have $Z \sim N(0,1)$ independent of $X$ and so of $\text{signum}(X)$. Since $Y= \rho X+\sqrt{1-\rho^2}Z$, you can then say  $$\text{Cor}(Y,\text{signum}(X)) \\=\text{Cov}(Y,\text{signum}(X)) \\= \text{Cov}(\rho X+\sqrt{1-\rho^2}Z,\text{signum}(X)) \\= \rho \text{Cov}(X,\text{signum}(X)) +\sqrt{1-\rho^2}\text{Cov}(Z,\text{signum}(X)) \\= \rho\sqrt{\tfrac{2}{\pi}}+\sqrt{1-\rho^2}0\\=\rho\sqrt{\tfrac{2}{\pi}}$$
