To determine if there exists a set $M$ in usual Euclidean space such that for every plane $\lambda$, the intersection $M \cap \lambda$ is finite and nonempty, we need to consider a construction that satisfies these conditions.

One possible approach is to construct the set $M$ using a version of the "space-filling curve" concept, but within certain constraints. However, space-filling curves like the Peano or Hilbert curves fill an entire region and would not make the intersection with a plane finite, thus a different approach is needed.

Instead, consider the following construction:

Construct the set $M$ by taking a dense set of points on every line parallel to one of the coordinate axes, such that these points are sparse in other coordinate directions. One way to do this is:

- For a subset of lines along the $x$-axis, $y$-axis, and $z$-axis (in $\mathbb{R}^3$), include points spaced in such a way that each point belongs to a single line only. Specifically, for each integer point on the $x$-axis of the form $(n, 0, 0)$, place a point $(n, \frac{1}{n}, \frac{1}{n})$.

This construction ensures:

1. Non-empty intersection: For any plane $\lambda$ in $\mathbb{R}^3$, there will be at least one axis-aligned line passing through or intersecting this plane at some point, and since we have points densely populating these lines, $M \cap \lambda$ is nonempty.

2. Finite intersection: Given the specific choice of constructing sparse points only on one type of line, the intersection of any plane $\lambda$ with these lines would result in a finite number of points on that plane.

Thus, the set $M$ satisfies the conditions of having finite and nonempty intersections with any plane $\lambda$.

Therefore, it is indeed possible to construct such a set $M$.

The existence of such a set $M$ in usual Euclidean space is conclusively:
$$\boxed{\text{yes}}$$