$$ \mathbb{E}[\operatorname{sign}(X)\operatorname{sign}(Y)]. $$
Let $Z_1, Z_2$ be independent standard normal variables and define a linear transformation so that $$ X = Z_1, \qquad Y = \rho Z_1 + \sqrt{1 - \rho^2}, Z_2. $$
Then $$ \mathbb{E}[\operatorname{sign}(X)\operatorname{sign}(Y)] = \mathbb{P}(X>0, Y>0) + \mathbb{P}(X<0, Y<0) - \mathbb{P}(X>0, Y<0) - \mathbb{P}(X<0, Y>0). $$
By symmetry, $$ \mathbb{P}(X>0, Y>0) = \mathbb{P}(X<0, Y<0), $$ and $$ \mathbb{P}(X>0, Y<0) = \mathbb{P}(X<0, Y>0). $$
Since the four quadrants partition the plane, $$ 2,\mathbb{P}(X>0, Y>0) + 2,\mathbb{P}(X>0, Y<0) = 1. $$ Therefore $$ \mathbb{E}[\operatorname{sign}(X)\operatorname{sign}(Y)] = 4,\mathbb{P}(X>0, Y>0) - 1. $$
Thus we only need $\mathbb{P}(X>0, Y>0)$.
From the representation of $Y$, $$ Y>0 \quad\Longleftrightarrow\quad \rho X + \sqrt{1-\rho^2}, Z_2 > 0 \quad\Longleftrightarrow\quad Z_2 > -\frac{\rho}{\sqrt{1-\rho^2}}, X. $$
Hence $$ \mathbb{P}(X>0, Y>0) = \mathbb{P}!\left( X>0,; Z_2 > -\frac{\rho}{\sqrt{1-\rho^2}}, X \right). $$
Let $$ \alpha = -\frac{\rho}{\sqrt{1-\rho^2}}. $$
We need $$ \mathbb{P}(x>0,\ z>\alpha x) = \iint_{x>0,\ z>\alpha x} \frac{1}{2\pi} \exp!\left(-\frac12(x^2 + z^2)\right), dz, dx. $$
Switch to polar coordinates $x = r\cos\theta$, $z = r\sin\theta$.
The line $z = \alpha x$ becomes $\tan\theta = \alpha$, i.e. $\theta = \arctan(\alpha)$.
Thus the region $x>0$ and $z>\alpha x$ corresponds to $$ \theta \in [\arctan(\alpha),, \tfrac{\pi}{2}], \qquad r>0. $$
Compute: $$ \mathbb{P}(X>0, Y>0) = \frac{1}{2\pi} \int_{\arctan(\alpha)}^{\pi/2} \int_0^\infty r, e^{-r^2/2}, dr, d\theta. $$
The $r$–integral is $$ \int_0^\infty r, e^{-r^2/2}, dr = 1. $$
Hence $$ \mathbb{P}(X>0, Y>0) = \frac{1}{2\pi}\left( \frac{\pi}{2} - \arctan(\alpha) \right). $$
Recall $\alpha = -\frac{\rho}{\sqrt{1-\rho^2}}$, so $$ \arctan(\alpha) = -\arctan!\left(\frac{\rho}{\sqrt{1-\rho^2}}\right) = -\arcsin(\rho). $$
Thus $$ \mathbb{P}(X>0, Y>0) = \frac{1}{2\pi}\left( \frac{\pi}{2} + \arcsin(\rho) \right). $$
Finally, $$ \mathbb{E}[\operatorname{sign}(X)\operatorname{sign}(Y)] = 4,\mathbb{P}(X>0, Y>0) - 1 = \frac{2}{\pi}\arcsin(\rho). $$