Rice's theorem vs Turing completeness I would like to clarify this because I see some kind of contradiction between Rice's theorem and Turing completeness.

This is the problem:

In building an Universal Turing Machine to emulate another Turing Machine, how can we be sure that we are indeed simulating it? ('sure' in the sense of being computationally verified by an algorithm). So I get confused, the problem I found is that Rice's theorem apparently tell us that we can not verify this computationally.. Is not "to emulate" an extensive property?.This lead me to think we could not prove the Universal Turing Machine is doing what it's supposed to do.. 

EDIT

To expand the topic

I want to expand this point, what I understood from given answers, is you can state that a specific Machine is Universal but not that an arbitrary machine is (nor a family of machines are universal), ok, but I don't ask for arbitrary!, nor a family, but to test algorithmically a single fixed machine. Test, like measuring, is against something, It's hard to think a computer concept without having something to compare with, and according to answers, completeness of a machine seems as something that can't be compared.

How universality is proven? let's say by showing 'look mom I am simulating a TM', so, as she believes TM are universal, then your machine is also universal, but, she can't verify it!, she must believe you because she can't test that single machine you made, not by an algorithm, so that's why I see Rice's theorem rubs against completeness, there is no algorithm to test two fixed machines are doing the same computation. 

So what can "universal" mean?, it's seems like saying, your software is doing what you mind, because is written in a Turing Complete language!. 'This is emulating a machine because it can'. I think that's not enough.
Your misunderstanding is:


  'sure' in the sense of being computationally verified by an algorithm


We are not, and we can not be . The question,


  Is this given Turing machine $M$ a universal one?


can not be generally and algorithmically decided for the reasons you state.

However, we can prove for a fixed Turing machine that it is universal -- and that is quite enough.