A knot is smoothly slice exactly when its zero-trace embeds in $S^4$

Suppose $D\subset B_1$ is a smooth proper slice disk for $K$. Its oriented normal $2$-plane bundle is trivial because $D$ is a disk. Choose a nowhere-zero normal section and push $D$ to a parallel disk $D^{\mathrm{\prime }}$ in the exterior $X$ of a smaller product tube of $D$. Its boundary $K^{\mathrm{\prime }}$ is a degree-one parallel of $K$, identified radially with a curve on the fixed boundary torus. It bounds $D^{\mathrm{\prime }}$ in $X$, so [F3] makes its class zero in $H_1(E_K)$. It is therefore a preferred longitude in the sense of [F2]. To compare with the fixed framing, let $s_0$ be its section pushoff representing ${\lambda }_K$. In product coordinates on the tube, the classes of $s_0$ and ${\mu }_K$ form a basis of the torus homology: its two coordinate circles generate $H_1(S^1\times S^1;\mathbb{Z})$. The disk-induced section has class $[K^{\mathrm{\prime }}]=[s_0]+n[{\mu }_K]$, where $n$ is the winding number of its normal direction relative to $s_0$. Both section classes vanish in $H_1(E_K)$, so $n=0$ by [F6]. The relative direction map $S^1\to S^1$ consequently has a periodic real-valued angle lift; scaling that lift to zero homotopes the two normal framings. Fibre rotations along this homotopy give isotopic attaching tubes. Thus the disk tube induces the fixed preferred-longitude framing, rather than merely some curve satisfying [F2].

Conversely, suppose $i:X_0(K)\hookrightarrow S^4$ is a smooth codimension-zero embedding. In the $0$-handle of $X_0(K)$ take the cone on $K$ from an interior point $p$, and join that cone disk to the core disk $D^2\times \{0\}$ of the $2$-handle along their common boundary $K$. After smoothing the join, this is a piecewise-smooth embedded $2$-sphere, smooth away from the cone point $p$.

Use oriented coordinates at $p$ and standard stereographic coordinates centred at $i(p)$. The derivative of $i$ is invertible by [F4], because its differential is injective and both dimensions are four. By [F5], $i$ is a local diffeomorphism near $p$. If its local orientation sign is negative, compose it with a fixed orientation-reversing reflection of the standard $S^4$; this preserves the existence of the embedding and makes the local sign positive. Thus its derivative $A$ lies in ${\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$.

A tubular neighborhood of $D$ is $D^2\times D^2$ and meets the other hemisphere $B_2$ in the attaching region $S^1\times D^2$. Consequently $B_2\cup \nu (D)$ is obtained from a $4$-ball by attaching one $0$-framed $2$-handle along $K$, hence is diffeomorphic to $X_0(K)$ by [F1]. This union is a smooth codimension-zero submanifold of $S^4$.

Identify the cone near $p$ with the radial cone on $K\subset S^3$. Along each direction $u\in K$, the derivative at the cone point of the target distance is $\parallel Au\parallel >0$. Compactness of $K$ makes this derivative uniformly positive, so [F5] gives a smooth radial intersection parameter for all sufficiently small target spheres. Taylor expansion in $C^1$ along $K$, followed by radial normalization, makes the resulting link ${\mathrm{\Gamma }}_{ϵ}$ $C^1$-close to ${\varphi }_A(K)$, where ${\varphi }_A(u)=Au\mathrm{/}\parallel Au\parallel$. The polar decomposition $A=QP$ joins $A$ to the identity through ${\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$: join the positive-definite $P$ to $I$ linearly and join $Q\in \mathrm{S}\mathrm{O}(4)$ to $I$ by a path in the connected group $\mathrm{S}\mathrm{O}(4)$. Hence ${\varphi }_A(K)$ is isotopic to $K$. The chosen normal-disk tube of $K$, transported by ${\varphi }_A$, is a tube of ${\varphi }_A(K)$. For sufficiently small $ϵ$, the tubular projection restricts to a local diffeomorphism from ${\mathrm{\Gamma }}_{ϵ}$ to ${\varphi }_A(K)$. It is homotopic to the identity because the curves are $C^1$-close, so it has degree one; a local diffeomorphism of circles with degree one is a diffeomorphism. Thus ${\mathrm{\Gamma }}_{ϵ}$ is the graph of a small normal section. Scaling that section to zero gives a smooth knot isotopy to ${\varphi }_A(K)$, so ${\mathrm{\Gamma }}_{ϵ}$ is isotopic to $K$.

Choose the round ball bounded by that small sphere in the fixed stereographic chart. Its complement $W=S^4\setminus \mathrm{int}(B_{ϵ})$ is a standard $4$-ball, as seen by inversion in the chart with its point at infinity. Shrink $ϵ$ so that the ball misses the compact part of the image sphere outside the cone chart. The image sphere is smooth away from the cone point; transversality makes its remainder after deleting the cone cap a smooth properly embedded disk in $W$ with product boundary collar and boundary ${\mathrm{\Gamma }}_{ϵ}$. Append the trace annulus of the isotopy in step 2.2 inside a boundary collar of $W$, with the isotopy stationary near its ends. The result is a smooth slice disk with boundary exactly $K$.

Steps 1.1, 2.1 construct an embedding from a slice disk, and steps 1.2, 1.3, 2.2, 3.1 construct a slice disk from an embedding. These are the two implications in the statement.