How does Turing's thesis imply the existence of a universal Turing machine? In the fifth edition of Boolos et al's Computability and Logic, Exercise 4.5 asks the following:

A universal Turing machine is a Turing machine $U$ such that for any other Turing machine $M_n$ and any $x$, the value of the two-place function computed by $U$ for arguments $n$ and $x$ is the same as the value of the one-place function computed by $M_n$ for argument $x$.
Show that if Turing's thesis is correct, then a universal Turing machine must exist.

Intuitively, the set of Turing machines is enumerable, then for any finite $n$ we can proceed to enumerate the machines as finite strings. Once we get to the required $n$, we run the Turing machine $M_n$ on the argument $x$. If $M_n$ halts in a standard position with some output, then we have found $u(n,x)$. If $M_n$ does not halt or halts in a nonstandard position, then that is the desired "value" (presumably, $u(n,x)$ is then undefined, but that should be all right since by "function" Boolos et al mean a total or partial function, unless otherwise specified). This is an informal list of instructions to compute $u(n,x)$, so $u$ is effectively computable, and so by Turing's thesis is also Turing-computable, whence $U$ must exist.
That seems trivial. But the exercise is the final one in a block of exercises, and the instructor's manual hints at using the Turing-uncomputable diagonal function, whose relevance I am not seeing. This is evidence towards me missing something. What am I missing?
 A: The argument alluded to in the question is sound. Specifically: Turing's thesis says that, if a human computer can perform a given algorithm, then there is a Turing machine to do it. A human computer can certainly simulate the computation of a Turing machine, given the code for the machine. Therefore, by Turing's thesis, there must be a universal Turing machine that can do the same thing. 
I don't think that the instructor's hint is germane to the question as stated. Perhaps a different question was intended? 
A: Turing's thesis asserts that every "effectively computable" function is computable by a Turing machine. It is not necessary to sharply define "effective computability" to get the converse implication. This amounts to showing that there is a partial enumeration $F$ (possibly nonterminating) of the effectively computable functions such that the value of $F$ on input $x$ for any arbitrary index $e \in \mathbb N$ is computably enumerable by an arbitrary machine $U$.
More precisely, $U(e,x) \iff  F_e(x)$, when $F$ is partial computable and $U \iff F$ is shorthand for $U$ is defined $\iff$ $F$ is also defined, taking into account their respective inputs.
To show that $U$ exists then one has to construct a diagonal function $\psi$ which enumerates the domain of $F_e(x)$ in strictly increasing order. The reason that the existence of such a $U$ follows from Turing thesis is that this universal machine is a "well-defined" intuitive algorithm by constructing a fixed point $F_e(x) = F_{\psi(x)}(x)$ when $\psi(e)$ is a total function (so it is defined for every input). 
Thus, no matter what partial enumeration of the "effectively computable" functions is used, there will surely exist a partial function $U$ such that any arbitrary index $e$ in the enumeration will witness some total function $\psi$, taking $F_e(x)$ to $F_{\psi(e)}(x)$ on all its domain as a fixed-point, given input $x$.
