[proofplan]
We rewrite the formula as a finite-dimensional duality statement for the compact convex set $\operatorname{conv} U_A(x) \subset [0,1]^n$. The upper bound follows by comparing each nonnegative linear functional of Euclidean norm at most one with the Euclidean norm on this convex set. For the reverse inequality, we choose a point of minimal Euclidean norm in the convex hull and use the first-order variational inequality for the squared norm on a convex set. If $A=\varnothing$, both sides are interpreted as $+\infty$ by the stated convention.
[/proofplan]
[step:Reduce the assertion to a compact convex subset of $[0,1]^n$]
Fix $x = (x_1,\dots,x_n) \in X$ and $A \subset X$. If $A=\varnothing$, then $U_A(x)=\varnothing$, hence $\operatorname{conv}U_A(x)=\varnothing$ and
\begin{align*}
d_T(x,A)=\inf \varnothing=+\infty.
\end{align*}
For every $\alpha \in [0,\infty)^n$ with $|\alpha|\le 1$, the set over which the inner infimum is taken is empty, so
\begin{align*}
\inf_{y\in \varnothing}\sum_{i=1}^n \alpha_i\mathbb{1}_{\{x_i\ne y_i\}}=+\infty.
\end{align*}
The supremum of the constant value $+\infty$ is $+\infty$, and the identity follows.
Assume from now on that $A\ne\varnothing$. Define
\begin{align*}
S := U_A(x)=\{m_x(y):y\in A\}\subset \{0,1\}^n
\end{align*}
and define
\begin{align*}
C := \operatorname{conv}S \subset \mathbb{R}^n.
\end{align*}
Since $\{0,1\}^n$ is finite, $S$ is finite and nonempty. Therefore $C$ is a nonempty compact convex subset of $\mathbb{R}^n$, and $C\subset [0,1]^n$. By definition,
\begin{align*}
d_T(x,A)=\inf_{u\in C}|u|.
\end{align*}
For every $\alpha \in [0,\infty)^n$, linearity gives
\begin{align*}
\inf_{y\in A}\sum_{i=1}^n \alpha_i\mathbb{1}_{\{x_i\ne y_i\}}=\inf_{s\in S}\alpha\cdot s.
\end{align*}
Because the map $u\mapsto \alpha\cdot u$ is affine on $C=\operatorname{conv}S$, its infimum over $C$ equals its infimum over the finite set $S$. Hence the desired formula is equivalent to
\begin{align*}
\inf_{u\in C}|u|=\sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}.
\end{align*}
[/step]
[step:Bound every admissible weighted distance by the convex distance]
Let $\alpha\in[0,\infty)^n$ satisfy $|\alpha|\le 1$. For every $u\in C$, the Euclidean inner-product estimate gives
\begin{align*}
\alpha\cdot u \le |\alpha|\,|u| \le |u|.
\end{align*}
Taking the infimum over $u\in C$ on the left and using the fact that the inequality holds for every $u\in C$, we obtain
\begin{align*}
\inf_{u\in C}\alpha\cdot u \le \inf_{u\in C}|u|=d_T(x,A).
\end{align*}
Taking the supremum over all $\alpha\in[0,\infty)^n$ with $|\alpha|\le 1$ gives
\begin{align*}
\sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}\le d_T(x,A).
\end{align*}
[guided]
We first prove that no admissible weight vector can produce a value larger than the convex distance. Fix a vector $\alpha\in[0,\infty)^n$ with $|\alpha|\le 1$. The linear functional
\begin{align*}
L_\alpha:\mathbb{R}^n&\to\mathbb{R}
\end{align*}
is defined by $L_\alpha(u)=\alpha\cdot u$. For each $u\in C$, the elementary Euclidean inner-product bound gives
\begin{align*}
L_\alpha(u)=\alpha\cdot u\le |\alpha|\,|u|.
\end{align*}
Since $|\alpha|\le 1$, this implies
\begin{align*}
\alpha\cdot u\le |u|.
\end{align*}
This inequality holds for every point $u$ in the convex hull $C$. Therefore the smallest value of the linear functional on $C$ cannot exceed the smallest Euclidean norm on $C$:
\begin{align*}
\inf_{u\in C}\alpha\cdot u\le \inf_{u\in C}|u|.
\end{align*}
By the definition of Talagrand's convex distance, the right-hand side is $d_T(x,A)$. Since the same estimate holds for every admissible $\alpha$, taking the supremum over all admissible $\alpha$ preserves the inequality:
\begin{align*}
\sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}\le d_T(x,A).
\end{align*}
[/guided]
[/step]
[step:Choose the nearest point in the convex hull]
The continuous map
\begin{align*}
N:C&\to[0,\infty)
\end{align*}
defined by $N(u)=|u|$ attains its minimum on the compact set $C$. Let $u_0\in C$ be a minimizer, so
\begin{align*}
|u_0|=\inf_{u\in C}|u|=d_T(x,A).
\end{align*}
Since $C\subset[0,1]^n$, we have $u_0\in[0,\infty)^n$.
If $u_0=0$, then $d_T(x,A)=0$. The upper bound from the previous step and the admissible choice $\alpha=0$ give
\begin{align*}
0\le \sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}\le 0,
\end{align*}
so equality holds.
Assume now that $u_0\ne 0$. Define
\begin{align*}
\alpha_0:=\frac{u_0}{|u_0|}\in[0,\infty)^n.
\end{align*}
Then $|\alpha_0|=1$, so $\alpha_0$ is admissible.
[/step]
[step:Use the variational inequality for the nearest point]
We prove that
\begin{align*}
u_0\cdot u\ge |u_0|^2
\end{align*}
for every $u\in C$. Fix $u\in C$ and define
\begin{align*}
\gamma_u:[0,1]&\to C
\end{align*}
by $\gamma_u(t)=(1-t)u_0+tu$. Convexity of $C$ implies that $\gamma_u(t)\in C$ for every $t\in[0,1]$. Since $u_0$ minimizes the squared norm on $C$, for every $t\in[0,1]$,
\begin{align*}
|u_0|^2\le |\gamma_u(t)|^2.
\end{align*}
Expanding the square gives
\begin{align*}
|\gamma_u(t)|^2=|u_0+t(u-u_0)|^2=|u_0|^2+2t\,u_0\cdot(u-u_0)+t^2|u-u_0|^2.
\end{align*}
Subtracting $|u_0|^2$ and dividing by $t>0$ yields
\begin{align*}
0\le 2u_0\cdot(u-u_0)+t|u-u_0|^2.
\end{align*}
Letting $t\downarrow 0$ gives
\begin{align*}
0\le u_0\cdot(u-u_0).
\end{align*}
Equivalently,
\begin{align*}
u_0\cdot u\ge |u_0|^2.
\end{align*}
Therefore, for the admissible vector $\alpha_0=u_0/|u_0|$,
\begin{align*}
\alpha_0\cdot u\ge |u_0|
\end{align*}
for every $u\in C$. Taking the infimum over $u\in C$ gives
\begin{align*}
\inf_{u\in C}\alpha_0\cdot u\ge |u_0|=d_T(x,A).
\end{align*}
[/step]
[step:Convert the supporting functional back to weighted mismatches]
Since $\alpha_0$ is one of the admissible vectors in $[0,\infty)^n$ with $|\alpha_0|\le 1$, the previous step gives
\begin{align*}
\sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}\ge d_T(x,A).
\end{align*}
Together with the upper bound already proved, this gives
\begin{align*}
d_T(x,A)=\sup\{\inf_{u\in C}\alpha\cdot u:\alpha\in[0,\infty)^n,\ |\alpha|\le 1\}.
\end{align*}
Finally, for each admissible $\alpha$,
\begin{align*}
\inf_{u\in C}\alpha\cdot u=\inf_{s\in S}\alpha\cdot s=\inf_{y\in A}\sum_{i=1}^n\alpha_i\mathbb{1}_{\{x_i\ne y_i\}}.
\end{align*}
Substituting this identity into the preceding display proves
\begin{align*}
d_T(x,A)=\sup\left\{\inf_{y \in A} \sum_{i=1}^n \alpha_i \mathbb{1}_{\{x_i \ne y_i\}} : \alpha \in [0,\infty)^n,\ |\alpha| \le 1\right\}.
\end{align*}
This is the asserted dual formula.
[/step]
