Think of a simple example:

is the xy plane, and

is the subspace consisting of all points on the line

You can show that it is a subspace, because it contains the origin, and is closed under scalar multiplication and vector addition.

Now then. If your vector

then obviously the best approximation is

itself, because then

If, on the other hand,

then what is the vector in

closest to

Well, it would be the point on the line that is closest to

And that point is the point you would get by dropping the perpendicular from

to the line

That also turns out to be the orthogonal projection onto the subspace

The essential idea here is that the straight line (in this case, the perpendicular), is the shortest path between two points, at least in the Euclidean norm. And that "perpendicular" is equivalent to the orthogonal projection onto the subspace.

Make sense? Does this help?