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Let $C=\Vert T\Vert$. We construct the extension by a deterministic recursion over the given dense list, using the one-dimensional dominated extension of a real linear functional. Neither Zorn's lemma nor any selection of a family of extensions is needed.

Suppose $f:W\to \mathbb{R}$ is already a linear functional satisfying $|f(w)|\le C\Vert w\Vert$, and $v\notin W$. An extension to $W+\mathbb{R}v$ must have the formDefineFor any $w,z\in W$, the bound on $f$ and the triangle inequality giveThus every quantity in the supremum defining $L_v$ is at most every quantity in the infimum defining $U_v$. Taking $w=0$ and $z=0$ gives finite bounds, specificallyThe completeness of the real numbers defines these endpoints uniquely. Set $a=L_v$; this is an explicit rule, not an arbitrary choice from a nonempty interval.

For $t>0$, the upper-endpoint inequality with $w/t$ yields $f(w)+ta\le C\Vert w+tv\Vert$. For $t<0$, write $t=-s$ with $s>0$ and use the lower-endpoint inequality with $w/s$ to obtain the same bound. For $t=0$ it is the original bound. Applying the upper bound also to the negative of the vector givesThe representation $w+tv$ is unique because $v\notin W$, so $f_v$ is a well-defined bounded linear functional extending $f$.

Start with $W_0=U$ and $f_0=T$. At step $n$, if $y_n\in W_{n-1}$ leave the functional unchanged; otherwise use exactly the supremum-endpoint rule with $v=y_n$. Put $W_n=U+\mathrm{span}\{y_1,\dots ,y_n\}$. This defines compatible functionals $f_n$ on $W_n$ with the common bound $C$. Their union $f_{\infty }$ on $W_{\infty }={\bigcup }_n{W}_n$ is linear and bounded by $C$. The subspace $W_{\infty }$ is dense in $V$ because it contains every $y_n$.

To make the final passage to the closure equally free of the axiom of choice, for $v\in V$ let $j_n(v)$ be the least positive integer with $\Vert y_{j_n(v)}-v\Vert <2^{-n}$. Density ensures that this integer exists, and leastness specifies it uniquely. DefineThe limit exists becauseAny other approximating sequence in $W_{\infty }$ gives the same limit, by the same norm bound. Applying this independence to sums and scalar multiples of approximating sequences proves linearity of $\tilde{T}$. It also shows $\tilde{T}=f_{\infty }$ on $W_{\infty }$ and $|\tilde{T}(v)|\le C\Vert v\Vert$. Restriction to $U$ therefore givesThe upper bound follows from construction and the lower bound from restriction to $U$. This also covers $C=0$, when all extensions are zero. No completeness of $V$ or closedness of $U$ was assumed. The argument establishes the choice-free Hahn-Banach extension in a separable space by uniquely specified operations at every stage.