Why do we get a contradiction for $x=n$? I am looking at the following proof that the halting problem is not recursive. 
We suppose that it is recursive, i.e., for the given universal turing machine $U$ and the produced enumeration of all the turing machines as $T_1, \dots , T_n$, there is a turing machine $M$ such that $M$ with input $(n,x)$ gives output $1$ if $T_n$ with input $x$ halts and output $0$ if $T_n$ with input $x$ does not halt. It is obvious that we can convert $M$ to a turing machine $M'$ with the following property: 
$M'$ with input $(n,x)$ halts iff $M$ with input $(n,x)$ halts with output $0$. 
We check what $M'$ does with input $(n,n)$ and we get a contradiction. 
We have the following: 
$M(n,x)=\left\{\begin{matrix} 
1 \ \ \text{ if } T_n(x)\downarrow \\ 
0 \ \ \text{ if } T_n(x)\uparrow 
\end{matrix}\right.$ 
and 
$M'(n,x)=\left\{\begin{matrix} 
\downarrow \ \ \text{ if } M(n,x)=0 \Leftrightarrow T_n(x)\uparrow \\ 
\uparrow \ \ \text{ if } M(n,x)=1 \Leftrightarrow T_n(x)\downarrow 
\end{matrix}\right.$ 
Why do we get a contradiction for $x=n$ ? 

For this to work properly, we need to define $M'$ as
$$M'(n) = \begin{cases} \downarrow \text{ if } M(n,n) = 0 \iff T_n(n) \uparrow \\
\uparrow \text{ if } M(n,n) = 1 \iff T_n(n) \downarrow\end{cases}$$
Now, since $M'$ is a Turing machine, it must have correspond to one of the Turing machines in our enumeration $T_1, T_2, \dots$, say $M' = T_m$. Then if we run $M'$ with input $m$ we get
$$M'(m) = \begin{cases} \downarrow \text{ if } T_m(m) = M'(m) \uparrow  \\ \uparrow \text{ if } T_m(m) = M'(m) \downarrow\end{cases}$$
Now we get a contradiction since $M'(m) \downarrow$ if $M'(m) \uparrow$ and $M'(m) \downarrow$ if $M'(m) \uparrow$.
