Why there is no Turing Machine that accepts the Diagonal Language?
In the following I assume that the matrix has an 1 in coordinates $(i,j)$ if $M_i$ accepts $\sigma_j$ and a $0$ otherwise (swapping all 1s and 0s does not affect the argument).
Consider the language $L = \{\sigma_i \in \Sigma^* : \sigma_i \not\in L(M_i) \}$ that contains all the words $\sigma_i$ such that the entry at coordinates $(i,i)$ of the matrix is $1$. I will first show that $L$ is not acceptable.
Suppose towards a contradiction that there is some Turing machine $M^*$ that accepts $L$.
Since the set of Turing machines is enumerable, and the rows of the matrix are associated with an enumeration of all Turing machines, there must be some integer $i \ge 1$ such that $M_i = M^*$.
Then consider the word $\sigma_i$ and suppose that the entry at coordinates $(i,i)$ is $1$.
From the way the matrix is constructed you know that $M_i$ accepts $\sigma_i$. However, by the definition of $L$, you also know that $\sigma_i \not\in L$. This implies that $M^* = M_i$ cannot accept $\sigma_i$. This is a contradiction.
If the entry at coordinates $(i,i)$ is $0$, then $M_i$ does not accept $\sigma_i$. However, $\sigma_i \in L$ implying that $M^* = M_i$ accepts $\sigma_i$. Again, this is a contradiction.
Going back to your language $L_d$: if $L_d$ were acceptable then so would be $L$.
Suppose towards a contradiction that there is a Turing Machine $M$ that accepts $L_d$.
To accept $L$ proceed as follows:

find the index $i$ such that $\sigma_i = x$. This can be done by enumerating the words in $\Sigma^*$.
Simulate $M$ with input $i$.
If $M$ with input $i$ halts and accepts, accept.

Since this is a contradiction, $L_d$ cannot be acceptable.