If any triangle has area at most 1 , points can be covered by a rectangle of area 2.

An even simpler counterexample would be the corner points of any non-rectangular parallelogram of area 2, e.g. $(0,0)$, $(0,1)$, $(2,2)$ and $(2,1)$. Any triangles formed from those points must have half the area of the parallelogram, i.e. 1, but as the parallelogram is not a rectangle, any rectangle that covers it must have an area greater than 2.

In the arrangement of four points below, each of the $4$ triangles has area $1$.

$\hspace{7mm}$

As $L\to\infty$, it seems that the smallest rectangle enclosing the $4$ points would have to be about $L\times4/L$, which would give it an area of $4$, not $2$.

If the theorem is true, a circle that is approximated by a polygon of sufficiently enough vertices verifies the theorem too.

The radius of the circle is $r$. The rectangle is a square and has area $4 r^2$.

The triangle of greatest area in a circle is equilateral, so has area $3/2 \times r \times \sqrt{3}/2 \times r$.

If the theorem is true, $4r^2 \leq 2 \times (3/2 \times r \times \sqrt{3}/2 \times r)$.

So $16 \leq 27 /2$. Contradiction.

The theorem seems to be false.

I believe that the theorem you stated is false, but there's a simple variation of your statement that you can actually prove!

revised theorem: Given a finite number, n, of points in a plane such that any three points form a triangle of area 1 or less, then there is a triangle of area 4 in the plane that contains all $n$ given points.

A quick outline of why this is true:

-We have finitely many points in the plane, and hence finitely many ways to form a triangle. So there is a least upper bound $\alpha$ on the areas of the triangles, and at least one triangle has area $\alpha$. Start by drawing this triangle, which vertices we will call A,B and C, and which side we will call (AB), (BC), and (CA).

-From triangle ABC, we can draw a bigger triangle and in so doing complete the proof. Draw the line (AB)' which is parallel to (AB) and passes through vertex C. Similarly draw (BC)' which is parallel to (BC) and passes through vertex A, and (CA)' which is parallel to (CA) and passes through vertex B.
The intersection of (AB)',(BC)', and (CA)', is a triangle of area $4*\alpha \leq 4 $, which we'll call ABC'.

-The important point is that, by construction, ABC' must contain all n of our original points. Indeed, pick a point D distinct from A, B, C. We could draw the triangles ABD,ACD, or BCD, each of which has area less than or equal to $\alpha$. Keeping in mind that $Area=\frac{bh}{2}$, we see that area(ABD)$\leq$area(ABC) implies that the shortest distance between D and (AB) must be less than or equal to the shortest distance between C and (AB). In other words, D must lie between (AB) and (AB)' or between (AB) and the reflection of (AB)' over (AB). Likewise, the shortest distances between D and (CA) and between D and (BC) must be respectively less than or equal to the shortest distances between B and (CA) and A and (BC).
So D must lie also between (CA) and (CA)' or (CA) and the reflection of (CA)' over (CA), and D must lie between (BC) and (BC)' or (BC) the reflection of (BC)' over (BC). This restriction on where the point D can lie can be shown quite easily to be equivalent to the restriction that D must lie in the triangle ABC'. So there is a triangle of area $4*\alpha$, which is less than or equal to 4 by assumption, which contains every point of our plane.

-Informally, this proves the revised version of the theorem you gave. Sorry it's a couple years late, hope it's still interesting or helpful for somebody.