A square with edge length $1$ has area $1$. An equilateral triangle with edge length $1$ has area $\sqrt{3}/4 \approx 0.433$. So three such triangles have area $\approx 1.3$, but it requires four such triangles to cover the unit square, e.g.:
Q. How can it be proved that three unit triangles cannot cover a unit square?
I am not seeing a straightforward route to proving this.
On
(This was intended to be a comment but was too long.)
Here is another approach, but a bit messy. We can calculate how much of the perimeter of the square an equilateral triangle can hold. There are two cases, one where a side of the triangle is a side of the square which gives that one equilateral triangle can hold only length 1. The other case is where a corner of the square is inside the triangle. WLOG, we can assume the corner touches a side of the triangle since moving the corner inside until it does only increases the length of the perimeter inside the triangle. After some calculations, we get that the length of the perimeter inside the triangle is at most $\frac{\sqrt{3}}2(1-x)+x$ where $0 < x < 1$ which has a maximum of $1$ also.
Suppose we define the corner-power of a triangle as follows:
A triangle gets $1$ corner-power point for each corner of the square that it contains excluding the vertices of the triangle.
A triangle gets $\frac{1}2$ of a corner-power point for each corner of the square that is also a vertex of the triangle.
It should be clear that, in order for a neighborhood of the square's corner to be covered by a set of triangles, it must either:
Be contained within a triangle, but not on the vertex.
Be contained as the vertex of two triangles.
Hence, a cover of the square must have at least $4$ corner-power points. However, each triangle can either have a corner in its interior (1 point) or two vertices on corners (1 point) for a maximum of $1$ point. Three triangles gets at most 3 corner-power points, which is not enough to cover a square.