How can a TM M' have a property P if it only accepts a single string x from language of P?
I think I got it. Here is Rice's Theorem (Rice Theorem and Turing machine
b ehavior prop erties):



I looked it up here: Decidability for the definition of a problem in the context of Turing Machine (slides 9 and 11):


  
  Formally, a problem is a language.
  Each string encodes some instance.
  The string is in the language if and only if the answer to this instance of the problem is “yes.”
  
  
  A problem is decidable if there is an algorithm to answer it.
  
  
  Recall: An “algorithm,” formally, is a TM that halts on all inputs,    accepted or not. 
  Put another way, “decidable problem” = “recursive    language.”
  
  
  Otherwise, the problem is undecidable


That answers why I only need one string w and reduced x from w to prove a language L with property P is undecidable. 

According to Rice's Theorem, if language L with property P has Turing Machines that does not accept language of L, L is undecidable. Thus:

$$L_{p} = \{ L(M_{0}), L(M_{1}), L(M_{2}),..., L(M_{n})\}$$
$$L'_{NotInp} = \{ L(M_{0}), L(M_{1}), L(M_{2}),..., L(M_{n}) \}$$

$$M_{i}\mspace{8mu} \text{is}\mspace{8mu} \text{an}\mspace{8mu} i\text{th}\mspace{8mu} \text{Turing} \mspace{8mu} \text{machine},\mspace{8mu} i ≤ n. $$

If the language L' (which contains TMs not in P) is not an empty set, the above L sub P is undecidable. Recall that language with property P is decidable if and only if it is trivial, which means either it contains every RE languages or not at all. Since if a language L with property P is decidable, every Turing Machines accept every string in L, so we don't need to test the acceptance of input to a given TM. 

Also, according to the definition of a problem in the context of Turing Machine, if we have the answer (an algorithm) to every instance of the language L sub P (obviously, since L sub P contains every TM), L sub P is decidable. Otherwise, it's not.

Let's be more concrete with an example from here: More Reduction Examples



Aside from the proof given on the page, let's look at it more intuitively. We may have Turing Machine M which accepts the three strings. Other than that, the input w can be anything which is not in M, but belong to other M'. Input w can be "Not UIUC", "Not Iowa", or "Not Michigan" or any other strings which do not belong to M. Since we do not have the answer to every instance of the problem with M (because if we do, M will accept every string since every string has property P. In this case, it's not). The language is not trivial, thus it is undecidable.

This is my reasoning. What do you think?