So, topologically, $dS_4 \simeq \mathbb{R} \times S^3$ and is given by the hyperboloid of one sheet in $\mathbb{R}^{4,1}$ as you've written (although I think you missed $x_4$ in your defining equation). The isometries of this surface are just the isometries of the ambient space which fix the origin $O$ of $\mathbb{R}^{4,1}$. And the geodesics on this surface are precisely the intersections, in the ambient space, between the surface and planes through $O$.

If two points on $dS_4$ are timelike- or null-separated, then I think the answer to your question is obvious. So let's consider when two points are spacelike-separated. Call the points $A$ and $B$, and consider the plane $OAB$.

Since $A$ and $B$ are spacelike-separated, there is an isometry of $\mathbb{R}^{4,1}$ which boosts and rotates such that the plane $OAB$ is mapped to $OA'B'$, where $A'$ and $B'$ lie in the $S^3$ at $x^0 = 0$.

Now we act with one more isometry. Take the midpoint $C'$ of the line $A'B'$, and consider the ray $OC'$. Now act with a boost in $\mathbb{R}^{4,1}$ which leaves $OC'$ invariant. This moves $A' \to A''$ and $B' \to B''$, where now, say, $A''$ has coordinate $x^0 > 0$ and $B''$ has coordinate $x^0 < 0$.

Now it is simple to see a totally geodesic spacelike surface which separates $A''$ and $B''$: it is given by the $S^3$ at $x^0 = 0$. Since we have obtained this result acting only with isometries, it must be true that any spacelike-separated points can be separated by a totally geodesic spacelike surface.

So, topologically, $dS_3 \simeq \mathbb{R} \times S^2$ and is given by the hyperboloid of one sheet in $\mathbb{R}^{3,1}$ as you've written. The isometries of this surface are just the isometries of the ambient space which fix the origin $O$ of $\mathbb{R}^{3,1}$. And the geodesics on this surface are precisely the intersections, in the ambient space, between the surface and planes through $O$. Thus, for example, the $S^2$ at $x^0 = 0$ is a totally geodesic surface.

If two points on $dS_3$ are timelike- or null-separated, then I think the answer to your question is obvious. So let's consider when two points are spacelike-separated. Call the points $A$ and $B$, and consider the plane $OAB$.

Since $A$ and $B$ are spacelike-separated, there is an isometry of $\mathbb{R}^{3,1}$ which boosts and rotates such that the plane $OAB$ is mapped to $OA'B'$, where $A'$ and $B'$ lie in the $S^2$ at $x^0 = 0$.

Now we act with one more isometry. Take the midpoint $C'$ of the line $A'B'$, and consider the ray $OC'$. Now act with a boost in $\mathbb{R}^{3,1}$ which leaves $OC'$ invariant. This moves $A' \to A''$ and $B' \to B''$, where now, say, $A''$ has coordinate $x^0 > 0$ and $B''$ has coordinate $x^0 < 0$.

Now it is simple to see a totally geodesic spacelike surface which separates $A''$ and $B''$: it is given by the $S^2$ at $x^0 = 0$. Since we have obtained this result acting only with isometries, it must be true that any spacelike-separated points can be separated by a totally geodesic spacelike surface.

So, topologically, $dS_4 \simeq \mathbb{R} \times S^3$ and is given by the hyperboloid of one sheet in $\mathbb{R}^{4,1}$ as you've written. The isometries of this surface are just the isometries of the ambient space which fix the origin.

If two points on $dS_4$ are timelike- or null-separated, then I think the answer to your question is obvious. So let's consider when two points are spacelike-separated. Call the points $A$ and $B$, and consider the plane $OAB$, where $O$ is the origin of the ambient space $\mathbb{R}^{4,1}$.

Since $A$ and $B$ are spacelike-separated, there is an isometry of $\mathbb{R}^{4,1}$ which boosts and rotates such that the plane $OAB$ is mapped to $OA'B'$, where $A'$ and $B'$ lie in the $S^3$ at $x^0 = 0$.

Now we act with one more isometry. Take the midpoint $C'$ of the line $A'B'$, and consider the ray $OC'$. Now act with a boost in $\mathbb{R}^{4,1}$ which leaves $OC'$ invariant. This moves $A' \to A''$ and $B' \to B''$, where now, say, $A''$ has coordinate $x^0 > 0$ and $B''$ has coordinate $x^0 < 0$.

Now it is simple to see a totally geodesic spacelike surface which separates $A''$ and $B''$: it is given by the $S^3$ at $x^0 = 0$. Since we have obtained this result acting only with isometries, it must be true that any spacelike-separated points can be separated by a totally geodesic spacelike surface.

So, topologically, $dS_4 \simeq \mathbb{R} \times S^3$ and is given by the hyperboloid of one sheet in $\mathbb{R}^{4,1}$ as you've written (although I think you missed $x_4$ in your defining equation). The isometries of this surface are just the isometries of the ambient space which fix the origin $O$ of $\mathbb{R}^{4,1}$. And the geodesics on this surface are precisely the intersections, in the ambient space, between the surface and planes through $O$.

If two points on $dS_4$ are timelike- or null-separated, then I think the answer to your question is obvious. So let's consider when two points are spacelike-separated. Call the points $A$ and $B$, and consider the plane $OAB$.

Since $A$ and $B$ are spacelike-separated, there is an isometry of $\mathbb{R}^{4,1}$ which boosts and rotates such that the plane $OAB$ is mapped to $OA'B'$, where $A'$ and $B'$ lie in the $S^3$ at $x^0 = 0$.

Now we act with one more isometry. Take the midpoint $C'$ of the line $A'B'$, and consider the ray $OC'$. Now act with a boost in $\mathbb{R}^{4,1}$ which leaves $OC'$ invariant. This moves $A' \to A''$ and $B' \to B''$, where now, say, $A''$ has coordinate $x^0 > 0$ and $B''$ has coordinate $x^0 < 0$.

Now it is simple to see a totally geodesic spacelike surface which separates $A''$ and $B''$: it is given by the $S^3$ at $x^0 = 0$. Since we have obtained this result acting only with isometries, it must be true that any spacelike-separated points can be separated by a totally geodesic spacelike surface.