$\DeclareMathOperator{\KK}{KK}$
$\DeclareMathOperator{\Ob}{Ob}$
$\DeclareMathOperator{\C}{\mathcal{C}}$
$\DeclareMathOperator{\top}{top}$
$\DeclareMathOperator{\id}{id}$
$\DeclareMathOperator{\pr}{pr}$
$\DeclareMathOperator{\N}{N}$
$\DeclareMathOperator{\HKK}{\mathbf{KK}}$
$\DeclareMathOperator{\const}{const}$
$\DeclareMathOperator{\Hom}{Hom}$
Rasmus already answered the second part of your question. Let us construct a
stable quasi-category and a triangulated equivalence from its homotopy category
to $\KK$. ($\Ob(\KK)$ shall be the class of separable C*-algebras.) I know a
different, more abstract, construction, but you have heard me describe it
before.
Let $𝐂^*$ be the set of separable C*-algebras of set-rank $ω⋅2$. Let $Δ$ be
the set of finite positive von Neumann ordinals.
For two C*-algebras $A$ and $B$ in $𝐂^*$ and $n + 1$ in $Δ$ let $𝐊^n(A, B)$ be
the set of all Kasparov cycles from $A$ to $\C(Δ^n_{\top}, B)$ – the
C*-algebra of continuous functions from the topological $n$–simplex to $B$.
Let $R$ be the strict $2$–category of sets, many-valued functions and
inclusions of functions.
Choose $n$ and $i$ in $Δ$, $γ\colon i → n$ order preserving, and $f$ in
$𝐊^n(A, B)$. Let $γ^*𝑓$ be the set of preimages of $γ^*[f]$ in $𝐊^i(A, B)$.
The map $n + 1 ↦ 𝐊^n(A, B)$ is thus contravariantly oplaxly functorial for
order preserving maps and many-valued functions. Let $𝐊(A, B)$ be that functor
– a simplicial $R$–set.
Let $(ϕ, ℰ, T)$ be in $𝐊^n(A, B)$. Let $ℱ$ be the $𝐙/2$–graded Hilbert
$\C(Δ^n_{\top}, B)$–module $\C(Δ^n_{\top}, ℰ)$ with the inner product $⟨ξ, η⟩(s)
≔ ⟨ξ(s), η(s)⟩(s)$ and the module structure $(ξ⋅b)(s) ≔ ξ(s)⋅b$. Define a
representation $ψ$ of $\C(Δ^n_{\top}, A)$ by even adjointable operators on $ℱ$ by
$ψ(a)(ξ)(s) ≔ ϕ(a(s))(ξ(s))$ and define an odd adjointable operator $S$ on $ℱ$
by $S(ξ)(s) ≔ T(ξ(s))$. Put $M(ϕ, ℰ, T) ≔ (ψ, ℱ, S)$.
For three C*-algebras $A$, $B$ and $C$ in $𝐂^*$ and $n + 1$ in $Δ$ let
$μ^n(A, B, C)$ be the many-valued function from $𝐊^n(B, C) × 𝐊^n(A, B)$ to
$𝐊^n(A, C)$ that associates $h ∈ 𝐊^n(A, C)$ with $(f, g) ∈
𝐊^n(B, C) × 𝐊^n(A, B)$ if $[h] = [M(f)] ∘ [g] ∈ \KK(A, \C(Δ^n_{\top}, C))$.
Choose $A ∈ 𝐂^*$ and $n ∈ Δ$. Consider $\C(Δ^n_{\top}, A)$ as a $𝐙/2$–graded
Hilbert module over itself in the usual way and let $ρ$ be the representation
of $A$ by the even operators of pointwise left-multiplication on
$\C(Δ^n_{\top}, A)$. We write $η^n(A)$ for the many-valued function defined on
$\{0\}$ that takes as values all the Kasparov cycles equivalent to
$(ρ, \C(Δ^n_{\top}, A), 0) ∈ 𝐊^n(A, A)$.
Convince yourself of the following for all $A, B, C, D ∈ 𝐂^*$, $n + 1, i + 1 ∈
Δ$, $γ\colon i + 1 → i + 1$ order preserving:
$∃f ∈ 𝐊^n(B, D): ((f, g), h) ∈ μ^n(A, B, D) ∧ ((d, e), f) ∈ μ^n(B, C, D) ⇔
∃j ∈ 𝐊^n(A, C): ((d, j), h) ∈ μ^n(A, C, D) ∧ ((e, g), j) ∈ μ^n(A, B, C);$
$μ^i(A, B, C) ∘ (γ^* × γ^*) = γ^* ∘ μ^n(A, B, C):
𝐊^n(B, C) × 𝐊^n(A, B) → 𝐊^i(A, B);$
$γ^* ∘ η^n(A) = η^i(A)$;
$μ^n(A, A, B)(\id × η^n(A)) ⊇ \pr_1 ∧ μ^n(A, B, B)(η^n(B) × \id) ⊇ \pr_2.$
Thus $(𝐂^*, 𝐊, μ, η)$ is a category enriched in the $2$–category of simplicial
$R$–sets. Let us call such objects simplicial $R$–categories, as categories
enriched in simplicial sets are – also in this answer – called simplicial
categories.
Let us write $Δ^n$ for the $n$–dimensional simplicial ($R$–)simplex, $Λ^n(k)$
for its $k$–th horn and $ι(n, k)$ for the inclusion of $Λ^n(k)$ into $Δ^n$. We
say that a simplicial $R$–set $X$ is a Kan complex if and only if for every
$n + 1$ and $k + 1$ in $Δ$ and every single-valued map $σ\colon Λ^n(k) → X$
there is a map $τ\colon Δ^n → X$ such that $σ = τ ∘ ι(n, k)$.
Claim: The simplicial $R$–set $𝐊(A, B)$ is a Kan complex for all $A$ and
$B$ in $𝐂^*$.
Proof. Pick any $n + 1$ and $k + 1$ in $Δ$. Every single-valued map
$σ\colon Λ^n(k) → 𝐊(A, B)$ determines some $f ∈ 𝐊^0(A, \C(Λ^n(k)_{\top}, B))$,
where $Λ^n(k)_{\top}$ is the corresponding $(n − 1)$–dimensional topological
horn. The map $B ↦ 𝐊^0(A, B)$ is functorial for ${}^*$–homomorphisms and
many-valued functions; thus $f$ lifts to some $g ∈ 𝐊^0(A, \C(Δ^n_{\top}, B))$.
Use $g$ to fill in the missing values and thereby extend $σ$ to $Δ^n$.
Let $ℭ(Δ^n)$ be the canonical realisation of Δⁿ as cofibrant simplicial
category.
Let $D$ be a simplicial $R$–category. We define the homotopy coherent nerve
$\N(D)$ of $D$ by
$$
\N(D)(n + 1) ≔ \Hom_{\operatorname{single-valued}}(ℭ(Δ^n), D).
$$
Claim: The homotopy-coherent nerve of a simplicial $R$–category $D$ is a
quasi-category if all the simplicial arrow $R$–sets are Kan complexes.
Proof. Choose $n ≥ 2$ and $i$ with $1 ≤ i ≤ n − 1$. Put $S ≔ ℭ(Δ^n)$ and
let $Q_k$ be the set of degenerate $k$–simplices in $S(0, n)$. Let $L$ be the
largest simplicial subcategory of $S$ with $\Ob(L) = \Ob(S)$ and
$$
∀k ≥ 1\colon
\{σ ∈ L_k(0, n) ∣ σ_0 = \{0, n\} ∧ σ_k ∪ \{i\} = \{0, \dotsc, n\}\} ⊆ Q_k.
$$
$L$ realises the horn $Λ^n(i)$ as cofibrant simplicial category.
Let $F\colon L → D$ be a functor. We need to extend $F$ to $S$. Because
$L(j, k)$ and $S(j, k)$ only differ for $j = 0$ and $k = n$ we only need to
extend the map $F(0; n)\colon L(0, n) → D(F(0), F(n))$ to $S(0, n)$. The
inclusion of $L(0, n)$ into $S(0, n)$ is a cofibration. It is also a weak
homotopy equivalence, because the geometric realisation of $S(0, n)$ is
homeomorphic to $[0, 1]^{n − 1}$, and the image of $L(0, n)$ is the contractible
subset
$$
\{x ∈ [0, 1]^{n − 1} ∣ x_i = 1 ∨ ∃j ≠ i: (x_j = 0 ∨ x_j = 1)\}
$$
under this homeomorphism. Therefore it is anodyne. It has the extension
property with respect to all simplicial $R$–sets that are Kan complexes because
the inclusion of simplicial sets into simplicial $R$–sets and single-valued
maps preserves colimits.
We now know that $\HKK ≔ \N(𝐂^*, 𝐊, μ, η)$ is a quasi-category. Its objects
are elements $A$ of $𝐂^*$ paired with a choice of one element from $η^n(A)$ for
every $n + 1$ in $Δ$. We can choose such elements canonically (see above); we
write $ι$ for this inclusion of $𝐂^*$ into $\Ob(\HKK)$. The restriction of
$\HKK$ to $𝐂^*$ is equivalent to $\HKK$ via the inclusion. On the other hand
also $\KK$ restricted to $𝐂^*$ is equivalent to $\KK$ via the inclusion. For
$A$ and $B$ in $𝐂^*$, the cycles for $π(\HKK)(ι(A), ι(B))$ map onto
$\KK(A, B)$. That map $θ$ is a bijection modulo equivalence because $θ(f) =
θ(g)$ in $\KK(A, B)$ only if $[M(h)] ∘ [f((s_0)^n(\const \{0, 1\}))] =
[g((s_0)^n(\const \{0, 1\}))]$ in $\KK(A, \C(Δ^n_{\top}, B))$ for all $n + 1$
in $Δ$ and some $h$ in $η^n(B)$.
We now only need to see that $\HKK$ is stable in a way that is compatible with
the triangulation on $\KK$.
For $0 ∈ 𝐂^*$, $ι(0)$ is a final object of $\HKK$, because $0$ is a final
object of $\KK$.
Claim: For all matrix-stable $A, B, D$ in $𝐂^*$, $f = (ϕ, D, 0)$ in
$𝐊^0(A, D)$ and $g = (ψ, D, 0)$ in $𝐊^0(B, D)$, a map $F: Δ^1 × Δ^1 → \HKK$ is
a Cartesian square if
$F_0(0; 0)
= \{(a, b, d, h_1, h_2) ∈ A ⊕ B ⊕ D ⊕ \C([0, 1], D) ⊕ \C([0, 1], D)
∣ ∀d_2 ∈ D: h_1(0)⋅d_2 = ϕ(a)(d_2)
∧ h_1(1) = d = h_2(1)
∧ h_2(0)⋅d_2 = ψ(b)(d₂)\}$,
- $F_0(0; 1) = A$,
- $F_0(1; 0) = B$,
- $F_0(1; 1) = D$;
- $F_1(0, 0; 0, 1) = ((a, b, d, h_1, h_2) ↦ a_2 ↦ aa_2, A, 0)$,
- $F_1(0, 1; 0, 0) = ((a, b, d, h_1, h_2) ↦ b_2 ↦ bb_2, B, 0)$,
- $F_1(0, 1; 1, 1) = f$,
- $F_1(1, 1; 0, 1) = g$,
- $F_1(0, 1; 0, 1) = ((a, b, d, h_1, h_2) ↦ d_2 ↦ dd_2, D, 0)$;
- $F_2(0, 0, 1; 0, 1, 1)
= ((a, b, d, h_1, h_2) ↦ ξ ↦ h_1ξ, \C([0, 1], D), 0)$,
- $F_2(0, 1, 1; 0, 0, 1)
= ((a, b, d, h_1, h_2) ↦ ξ ↦ h_2ξ, \C([0, 1], D), 0)$;
Remark: Every cospan in $\HKK$ is equivalent to one as in the claim.
Proof. Choose $n ≥ 1$ and a map $G\colon ∂Δ^n × Δ^1 × Δ^1 → \HKK$ of the
appropriate kind. Without loss of generality we can adjust your choice of $G$
so that all the Kasparov cycles in the image come from ${}^*$–homomorphisms
(because $F_0(0; 0)$ is matrix-stable). $G$ extends to $Δ^n × Δ^1 × Δ^1$
because $F$ determines a Cartesian square in the homotopy coherent nerve of the
topologically enriched category with $𝐂^*$ as set of objects and
${}^*$–homomorphisms with the topology of pointwise norm-convergence as mapping
spaces.
Bott periodicity therefore tells us that a suspension functor on $\HKK$ is an
equivalence. Therefore $\HKK$ is stable. The description of Cartesian squares
makes it clear that the triangulation on the homotopy category of $\HKK$ is
compatible with the equivalence to $\KK|𝐂^*$.