Q:

There is no surjective, isotone map from $P$ to $$

An exercise of "Universal Algebra: Fundamentals and Selected Topics" of Clifford Bergman. 

(a)  Let  $P$  be  a  poset.   Show  that  there  is  no  surjective, 
  isotone  map  from  $P$  to  $<Dn(P), \subseteq>$.  (Hint:  suppose 
  $f$  were  such  a  map.   Consider  the  set  $B  =  \{x  \in  P  :
\exists(y)\ x\le y  \ \ \& \ y \notin  f(y)   \}$.)
(b)  Use  part  (a)  to  prove  Cantor’s  theorem:  for  any  set  X, 
  there  is  no  surjective  function  from  X  to  Sb(X). 

Here $Dn(P)$ is the set of downsets of $P$.
(b) follows from (a) if we see $X$ as a poset with the trivial partial order $x \le y$ iff $x=y$. With this order relation $Sb(X)=Dn(X)$ and every map from $X$ to $Sb(X)$ is isotone, hence there is no surjective map. 
It is intuitive, making some attempts, that (a) is true at least for finite sets.  I'm interested in a proof that uses the definition of $ B $, which I can't even prove to be non-empty. 

A:

The set B is nonempty. 
Reason: since f is surjective, something gets mapped to the empty set.   If f(x) = ∅, then x ∈ B since ∃y (e.g., y = x) with the given properties. 
Next observe that B is a downset, so f maps something in P to B.
Let b ∈ P be such that f(b) = B.
If b ∉ B, then there is no y ≥ b such that y ∉ f(y). 
But what about b itself?  Indeed, we have b ≥ b and b ∉ B = f(b), so b satisfies the criteria for membership in B (contradiction). 
Since assuming b ∉ B leads to a contradiction, assume b ∈ B.
Then there exists y ≥ b such that y ∉ f(y).  Let f(y) = C (so y ∉ C). 
Now use monotonicity to obtain the contradiction y ∈ C.

