The finite wiretap auxiliary has cardinality at most |X|

If $m:=\mathrm{\mid }\mathcal{X}\mathrm{\mid }=1$, then $X$ is constant and both mutual informations in the objective are zero, so $\mathrm{\mid }\tilde{\mathcal{U}}\mathrm{\mid }=1$ suffices. Assume $m\ge 2$. For a probability vector $q$ on $\mathcal{X}$, let $F(q):=I_q(X;Y)-I_q(X;Z)\mathrm{.}$ By [F2]–[F4] and the finite entropy identity in [F3], $F(q)=H(q{W}_Y)-H(q{W}_Z)-{\sum }_{x}q(x)(H(W_Y(\cdot \mid x))-H(W_Z(\cdot \mid x)))\mathrm{.}$ The maps $q\mapsto q{W}_Y$ and $q\mapsto q{W}_Z$ are finite linear maps. On $0<t\le 1$, set $s=t^{-1\mathrm{/}2}\ge 1$; [F5] gives $0\le -t\ln t=2t\ln s\le 2\sqrt{t}$. Hence $-t\ln t$ extends continuously to $t=0$ with value $0$, so finite entropy and $F$ are continuous on the simplex $\mathrm{\Delta }(\mathcal{X})$.

Write ${\lambda }_u=P_U(u)>0$, $q_u=P_{X\mid U=u}$, and $\bar{q}=P_X={\sum }_u{\lambda }_u{q}_u$, omitting zero-mass $u$. The chain rule [F3] gives $I(U;Y)-I(U;Z)=F(\bar{q})-{\sum }_u{\lambda }_{u}F(q_u)\mathrm{.}$ Indeed, expanding the right side by the entropy formula in step 1.1 leaves $H(Y)-{\sum }_u{\lambda }_{u}H(Y\mid U=u)-H(Z)+{\sum }_u{\lambda }_{u}H(Z\mid U=u)$; the linear conditional-on-$X$ terms cancel because $\bar{q}={\sum }_u{\lambda }_u{q}_u$, and the remaining expression is the left side.

Suppose the current posterior mixture has $k>m$ positive weights. The $k$ vectors $(q_u(x_1),\dots ,q_u(x_{m-1}),1)$ lie in ${\mathbb{R}}^m$, so there are coefficients $c_u$, not all zero, with ${\sum }_u{c}_u=0$ and ${\sum }_u{c}_u{q}_u(x_j)=0$ for $j<m$. Since each $q_u$ sums to one, the same relation holds for the last coordinate $x_m$. The coefficients include both signs. Therefore the weights ${\lambda }_u+t{c}_u$ remain nonnegative on a bounded interval $[t_{-},t_{+}]$ with $t_{-}<0<t_{+}$; at either endpoint at least one weight is zero, while the total weight and posterior average stay fixed. The average $G(t)={\sum }_u({\lambda }_u+t{c}_u)F(q_u)$ is affine in $t$, so at least one endpoint has $G(t)\le G(0)$: otherwise zero, a convex combination of $t_{-}$ and $t_{+}$, would have a smaller value than both endpoints. Choose that endpoint and delete its zero weights. This keeps $\bar{q}$ fixed and does not decrease the wiretap objective in step 2.1.

Repeating step 3.1 finitely many times reduces any finite auxiliary to at most $m$ positive weights, preserving $P_X$ and not decreasing $I(U;Y)-I(U;Z)$. Thus every finite auxiliary is dominated by one with $\mathrm{\mid }\mathcal{U}\mathrm{\mid }\le m$.

The parameter space ${\mathrm{\Delta }}_m\times \mathrm{\Delta }(\mathcal{X}{)}^m$ of $m$ weights and $m$ posterior laws is a nonempty closed bounded subset of ${\mathbb{R}}^{m+m^2}$, since it is described by finitely many coordinate inequalities $0\le t\le 1$ and coordinate-sum equalities. It is compact by [F6]. By step 1.1 the objective $F(\bar{q})-{\sum }_{j=1}^m{\lambda }_{j}F(q_j)$ is continuous there, including zero weights, so it attains a maximum by [F7]. Step 4.1 shows this maximum equals the supremum over all finite auxiliaries.