The boundary of a nontrivial knot exterior is incompressible

Under the full-AC hypothesis [F1], the supplier in [F2] applies. Since $c$ is essential on the torus, it is nonseparating: a separating simple curve on a genus-one surface cuts off a disk on one side and is null-homotopic. Cutting the torus along $c$ therefore gives a connected orientable surface with two boundary circles and Euler characteristic zero, hence an annulus. An embedded arc joining its two boundary circles closes after regluing to a loop meeting $c$ transversely once. The algebraic intersection is therefore $\pm 1$, so $[c]$ is primitive in $H_1(T_K;\mathbb{Z})$.

Because $c=\mathrm{\partial }D$, its image in $H_1(E_K;\mathbb{Z})$ is zero. By [F2], write $[c]=q[\lambda ]$ for an integer $q$. Since both $[c]$ and $[\lambda ]$ are primitive, $q=\pm 1$. Orient $c$ to make $[c]=[\lambda ]$. By [F3], $c$ is isotopic on $T_K$ to a chosen simple representative of the preferred longitude.

In product coordinates $N(K)\cong S^1\times D^2$, the curve $c$ has core degree one by [F2], so its torus class is $(1,k)$ for an integer $k$. The graph ${\gamma }_k(t)=(e^{it},e^{ikt})$ has that class. By [F3], choose a smooth isotopy $c_s(t)=(u_s(t),z_s(t))$ on the boundary torus from $c$ to ${\gamma }_k$, stationary near its final time after smooth reparametrization. This isotopy need not trace an embedded annulus on the torus itself. Instead choose a smooth strictly decreasing radius $r(s)$ from $1$ to $1\mathrm{/}2$ with nonzero derivative, and use $(s,t)\mapsto (u_s(t),r(s)z_s(t))$ in the tube. Distinct $s$ have distinct radii, while each fixed-$s$ curve is embedded; moreover the time derivative has nonzero radial component and is independent of its circle tangent. This gives a smooth embedded annulus with outer boundary exactly $c$ and inner boundary the radius-$1\mathrm{/}2$ copy of ${\gamma }_k$. Continue it inward by $(r,t)\mapsto (e^{it},r{e}^{ikt})$ for $0\le r\le 1\mathrm{/}2$. The stationary final curve makes the collars agree; the inward annulus is smooth and regular even at $r=0$ and has core boundary $K$. The resulting annulus lies otherwise in the interior of $N(K)$.

If the given disk is only boundary-preimage embedded, first replace it using [F5] by a neat smooth disk in $E_K$ with the same boundary $c$. The annulus of step 3.1 has nonzero inward radial derivative at $c$ on the tube side. The resulting proper disk $D$ on the exterior side and this annulus on the tube side meet only along $c$. Their union is an embedded disk with boundary $K$; smoothing the corner along $c$ in a product collar leaves that boundary fixed. Thus $K$ bounds a smooth disk, which is a genus-zero Seifert surface. Full AC supplies the countable-choice hypothesis of [F4], so [F4] makes $K$ ambient isotopic to the unknot, contrary to the hypothesis. Therefore no such disk $D$ exists.