Very close! Here's one that works:
scala> def myFlatten[T](list: List[List[T]]): List[T] = for (xs <- list; x <- xs) yield x 
myFlatten: [T](list: List[List[T]])List[T]
Or use the built-in flatten
scala> List(List(1, 2), List(3)).flatten
res0: List[Int] = List(1, 2, 3)
scala> List(Set(1, 2), Set(3)).flatten  
res1: List[Int] = List(1, 2, 3)
It's instructive to see how to write this function without the for syntactic sugar.
scala> def myFlatten[T](list: List[List[T]]): List[T] = list flatMap identity
myFlatten: [T](list: List[List[T]])List[T]
scala> myFlatten(List(List(1, 2), List(3)))
res3: List[Int] = List(1, 2, 3)
UPDATE
BTW, the fact that List[List[T]] can be flattened to List[T] is 50% of the reason that List is a Monad. Generally, this is known as join. The other 50% comes from the fact you can map a function A => B across a List[A] to result in a List[B]. The general name for this is a Functor map. fmap and join on Wikipedia.
A different way of defining a Monad for type constructor M is with a pure operation, that takes a value of type A, and returns a M[A]; and a bind operation that takes an M[A], a function A => M[B], and results in M[B]. For Lists, pure == List(_), and bind = (l: List[A], f: (A => List[B])) => l.flatMap(f)