Can you make a sphere out of a plane?

Although you can't make a sphere from a plane, there are map projections that tessellate "naturally" (and place the tricky singular points in the ocean where people tend not to notice them). You can't, for topological reasons, avoid the points at the corners, but this kind of map does avoid some of the problems of mirroring and is continuous except at those corner points.

Most well known is the "Peirce quincuncial" projection. Wikipedia has an image showing the projection

^{Image by Strebe - Own work, CC BY-SA 3.0}

What you want to do is not possible because there is no flat sphere. That is, there is no way to put a metric on a topological sphere such that the curvature is everywhere zero. This can be shown using the Gauss-Bonnet theorem: the global curvature (by which I mean the integral of the curvature on the whole sphere) is equal to ($2\pi$ times) the Euler characteristic, which for a sphere is $2$ (and not $0$).

On the other hand, it is very well-known to gamers that there are flat tori: you just teleport on the other side when you hit a wall. This is illustrated by the fact that the Euler characteristic of a torus is $0$, so there can be a flat metric on a torus (and indeed you can define one by expressing the torus as a quotient of the plane).

A mathematical model

Assume you managed to trick the player into thinking they are on a sphere while they are really walking on an infinite plane. What would the world have to look like?

First of all, whenever the player is standing at some point $x$ on the flat world, they are deceived to think they really are at some point $i(x)$ on the imaginary spherical world. In other words, the player's imagination creates a mapping $i : \mathbb{R}^2 \to S^2$.

The assumption

As another answer points out, it is impossible for $i$ to be a local isometry because of the difference in curvatures of the plane and the sphere. Another easy argument is that on the sphere there is a triangle with three right angles, while on the plane clearly there is not. But we can relax our expectations and only demand that $i$ be a rough local isometry. What do I mean by that?

Our player is just a human and as such, they can't really distinguish between $1$ meter and $99$ centimeters, they also can't see very far away. Thus we assume that for each sufficiently close points $x, y \in \mathbb{R}^2$ the following equality up to a small margin $\varepsilon$ between distances on the plane and on the sphere holds: $$(1-\varepsilon) \cdot d_{\mathbb{R}^2}(x, y) \leqslant d_{S^2} \big( i(x), i(y) \big) \leqslant (1+\varepsilon) \cdot d_{\mathbb{R}^2}(x, y).$$

A solution

It can be proven (though it's quite technical) that under this assumption $e : \mathbb{R}^2 \to S^2$ must be a covering map. But $S^2$ is simply connected, so it follows that $\mathbb{R}^2$ is homeomorphic to $S^2$, which is a contradiction. Hence a function with the properties stated above does not exist.

Which means what you are trying to do - is impossible.