how does Kleene-Post show two languages that are not Turing reducible to each other?
Unfortunately, I don't possess a copy of Sipser, so I will just define all my notation. Let $T_0,T_1,\ldots$ an enumeration of all oracle Turing machines whose input is a word over some alphabet $\Sigma$. I will denote by $T_i^O(x)$ the output of the execution of $T_i$ on input $x$ with oracle $O$, or $\bot$ if the machine doesn't halt. We say that $T_i$ (Turing-)reduces $A$ to $B$ if $T_i^B(x)$ halts for all $x$, and returns the truth value of &quot;$x \in A$&quot;. We assume for simplicity that the output of $T_i^B(x)$ can always be interpreted as a truth value.
We will construct two sequences $A_0,A_1,\ldots$ and $B_0,B_1,\ldots$ of &quot;partial sets&quot;, that is, mappings from $\Sigma^*$ to $\{0,1,\bot\}$. Initially, $A_0$ and $B_0$ map all strings in $\Sigma^*$ to $\bot$. Moreover, $A_{i+1}$ is an extension of $A_i$, that is, if $A_i(x)\neq \bot$ then $A_{i+1}(x) = A_i(x)$. Furthermore, each $A_i$ or $B_i$ is defined (not equal to $\bot$) for only finitely many words in $\Sigma^*$.
We will eventually take $A$ to be a set extending all $A_i$, that is, if $A_i(x) = 0$ for some $i$ then $i \notin A$; if $A_i(x) = 1$ for some $i$ then $i \in A$; and otherwise the status of $i$ doesn't matter (for definiteness, let $i \notin A$). The extension property guarantees that $A$ is well-defined. We define $B$ similarly.
We construct $A_{2i+1},B_{2i+1}$ from $A_{2i},B_{2i}$ in a way which rules out $T_i$ being a reduction from $A$ to $B$. By assumption, there exists $x \in \Sigma^*$ such that $A_{2i}(x) = \bot$. Consider what happens when we try to run $T_i$ on input $x$ with oracle $B_{2i}$. If $T_i$ tries to apply the oracle on a word $y$ on which $B_{2i}$ is defined, then it gets $B_{2i}(y)$. If $B_{2i}(y) = \bot$, then we simulate both branches in parallel (each of them could further split down the road). One of the following must happen:

There is a branch on which computation halts.
All branches result in non-halting computations.

The second case is easy – $T_i$ cannot be a reduction from $A$ to $B$, since it doesn't halt on input $x$, whatever happens in later stages of the construction. So we take $A_{2i+1} = A_{2i}$ and $B_{2i+1} = B_{2i}$.
In the first case, we actually have to do something. Pick a halting branch. The branch corresponds to a choice of definite values for some strings in $\Sigma^*$ on which $B_{2i}$ is undefined. We form $B_{2i+1}$ from $B_{2i}$ by keeping all values already defined in $B_{2i}$, and defining the additional strings occurring in the halting branch according to the branch. We form $A_{2i+1}$ from $A_{2i}$ by keeping all values already defined in $A_{2i}$, and defining $A_{2i+1}$ on $x$ in the opposite way to what $T_i$ answers in the halting branch: if $T_i$ outputs Yes then $A_{2i+1}(x) = 0$, and if $T_i$ outputs No then $A_{2i+1}(x) = 1$. This ensures that $T_i$ does not reduce $A$ to $B$, since it outputs the wrong value on $x$.
We construct $A_{2i+2},B_{2i+2}$ from $A_{2i+1},B_{2i+1}$ in a similar way, switching the roles of $A$ and $B$, ruling out $T_i$ being a reduction from $B$ to $A$.
Since the final $A$ and $B$ &quot;encompass&quot; all $A_i,B_i$, by construction no $T_i$ reduces $A$ to $B$ or $B$ to $A$, and so there is no Turing reduction from $A$ to $B$ or from $B$ to $A$.

The dichotomy above (some branch halts / all branches never halt) is not computable, in the sense that given $A_{2i},B_{2i},T_i$ we cannot determine which of the two options occurs. But it is possible using an oracle to the halting problem, since we can construct a Turing machine which tries out all branches in parallel, and immediately halts if one of them halts. Furthermore, by &quot;going down the tree&quot; we can find a halting branch if such exists. This shows that we can compute the sequences $A_i,B_i$ using an oracle to the halting problem.
Now suppose that when choosing a string $x$, then we always choose the first such string on which $A_{2i}$ (or $B_{2i+1}$) is undefined, according to some fixed ordering. This guarantees that any string will be eventually defined by some $A_i$ and by some $B_j$. Since the sequences $A_i,B_j$ are computable using an oracle to the halting problem, this shows that the sets $A,B$ are computable using an oracle to the halting problem.

Here is an alternative proof. Let $\mu$ be an arbitrary measure on languages such that $\mu(L) = 0$ for any specific language $L$ (we say that $\mu$ has no atoms). For example, $\mu$ could correspond to the experiment in which each word is put in the language with probability $1/2$ independently.
Let $\mathbf{A},\mathbf{B} \sim \mu$. The probability that $T_i$ reduces $\mathbf{A}$ to $\mathbf{B}$ is 0, since fixing the outcome of $\mathbf{B}$ to be $B$, there is at most one language $L$ which $T_i$ reduces to $B$, and by assumption $\mu(L) = 0$. Since there are only countably many machines $T_i$, it follows that the probability that $\mathbf{A}$ reduces to $\mathbf{B}$ is zero. Similarly, the probability that $\mathbf{B}$ reduces to $\mathbf{A}$ is 0. We conclude that almost surely (that is, with probability 1), $\mathbf{A}$ doesn't reduce to $\mathbf{B}$ and $\mathbf{B}$ doesn't reduce to $\mathbf{A}$. In particular, there exist realizations $A,B$ such that $A$ doesn't reduce to $B$ and $B$ doesn't reduce to $A$.