= Solution
Use the pair-design convention: each pair of distinct points must lie in exactly one <Steiner block> of size $k$. With this convention the printed endpoint $k=1$ is false for every $v\geq2$, since a singleton contains no pair. Thus the intended assertion requires $k\geq2$. For $k=2$, take all pairs as <Steiner blocks> for any $v\geq2$. We now give a finite-field construction for every fixed $k\geq3$, without assuming the desired design-existence theorem.
Put $m=\binom{k}{2}$ and take an odd <prime> power $q\equiv1\pmod{2m}$. By cyclicity of $\mathbb F_q^\times$, it has a <subgroup> $H$ of index $m$; its order $(q-1)/m$ is even, so $-1\in H$. We will find a $k$-set $A=\{a_1,\ldots,a_k\}$ such that its $m$ unordered pair differences occupy the $m$ distinct <cyclotomic classes of a finite field> modulo $H$. The following argument supplies that set for every sufficiently large compatible $q$.
First establish the required two-class intersection estimate by elementary <character> sums. Let $\chi$ be a <multiplicative character of a finite field> of order $m$, with $\chi(\gamma)=\zeta_m$ for a generator $\gamma$. Extend every power $\chi^j$, including the trivial <character>, by zero at zero. For $C_i=\gamma^iH$,
$$
\mathbf1_{C_i}(x)=\frac1m\sum_{j=0}^{m-1}\zeta_m^{-ij}\chi^j(x).
$$
For a nontrivial additive <character> $\psi$ of $\mathbb F_q$ and a nontrivial multiplicative <character> $\eta$, its <Gauss sum of a finite-field character> satisfies
$$
|G(\eta)|^2
=\sum_{t\ne0}\eta(t)\sum_{y\ne0}\psi((t-1)y)
=(q-1)-\sum_{t\ne0,1}\eta(t)=q.
$$
Here the inner sum equals $q-1$ for $t=1$ and $-1$ otherwise, by additive-character <orthogonality>; multiplicative-character <orthogonality> gives the last equality. Such a nontrivial additive <character> exists by identifying the additive <group> with a <vector space> over its <prime field> and choosing a nonzero <linear functional> into that <prime field>.
If $\eta,\xi,\eta\xi$ are nontrivial, substituting $x=uz$, $y=(1-u)z$ in the product of their Gauss sums proves
$$
G(\eta)G(\xi)=J(\eta,\xi)G(\eta\xi),\qquad
J(\eta,\xi)=\sum_u\eta(u)\xi(1-u).
$$
The terms with $x+y=0$ sum to zero because $\eta\xi$ is nontrivial. Hence $|J|=\sqrt q$. If exactly one <character> is trivial, subtracting the excluded zero and one terms gives absolute value one. If both are nontrivial but their product is trivial, the substitution $u/(1-u)$ gives $J(\eta,\eta^{-1})=-\eta(-1)$, again of absolute value one. If both are trivial, the sum is $q-2$. Expanding the two cyclotomic indicators and scaling a nonzero translate to one therefore gives, uniformly in $i,j$ and $t\ne0$,
$$
|C_i\cap(t+C_j)|=\frac{q}{m^2}+O_m(\sqrt q).
$$
All constants depend only on the fixed $m$.
To extend a prescribed difference pattern on $r$ coordinates, fix desired classes $D_1,\ldots,D_r$ for the new differences $x-a_i$ and set
$$
M(a)=\#\{x:x-a_i\in D_i\text{ for all }i\},\qquad a\in\mathbb F_q^r.
$$
Average over all $q^r$ tuples, allowing repeated coordinates just for this average. With $f=(q-1)/m$, counting choices for each $a_i$ independently gives
$$
\mathbb EM=q(f/q)^r=q/m^r+O_{m,r}(1).
$$
For the second moment, sum over the two potential extensions $x,y$. The $x=y$ terms contribute $qf^r/q^r$. For $x\ne y$, the choices of $a_i$ number $q/m^2+O_m(\sqrt q)$ each, by the preceding intersection estimate. Thus
$$
\mathbb EM^2=q^2/m^{2r}+O_{m,r}(q^{3/2}),\qquad
\operatorname{Var}M=O_{m,r}(q^{3/2}).
$$
The <Chebyshev inequality> shows that the proportion of tuples with $M(a)<q/(2m^r)$ is $O_{m,r}(q^{-1/2})$. Inductively, any fixed compatible difference pattern on $r$ distinct coordinates is realized by at least $c_rq^r$ tuples, with $c_r>0$ independent of sufficiently large $q$. The assertion starts with $r=1$. Almost all tuples are outside the exceptional set, so at least half the realizing tuples have at least $q/(2m^r)$ extensions. Each extension is distinct from the old coordinates because its differences belong to nonzero classes. This proves the assertion for $r+1$. This <cyclotomic pattern extension by a second-moment bound> avoids the false assumption that every partial choice can be extended.
Prescribe a different one of the $m$ <cosets> to each pair $a_j-a_i$, $i<j$. Since $-1\in H$, changing an orientation does not change its <coset>, so the prescriptions are compatible. The preceding induction supplies $A$.
Choose $S\subset H$ containing one element from each pair $\{h,-h\}$. Form the <Steiner blocks>
$$
\mathcal B=\{sA+t:s\in S,\ t\in\mathbb F_q\}.
$$
For two distinct points $x,y$, their difference identifies exactly one unordered pair $\{a_i,a_j\}$ of $A$ having the same $H$-coset. Exactly one of $(x-y)/(a_i-a_j)$ and its negative lies in $S$. Choose the orientation $a,b$ of that pair making $s=(x-y)/(a-b)\in S$, and then set $t=x-sa$. This yields $x=sa+t$, $y=sb+t$. Both the oriented base pair and $(s,t)$ are unique, so precisely one <Steiner block> contains the pair. In particular different <Steiner block> parameters cannot produce repeated <Steiner blocks>. Equivalently, the <Steiner blocks> $sA$ form a <difference family for a Steiner 2-design>, and translation gives the <cyclotomic construction of a Steiner 2-design>.
There are infinitely many suitable <field> orders: choose one odd <prime> $\ell$ not dividing $2m$, and take $q=\ell^{\varphi(2m)u}$ for <integers> $u\geq1$. Euler's congruence gives $q\equiv1\pmod{2m}$, and these orders become arbitrarily large. Consequently
$$
\boxed{\text{For every }k\geq2,\text{ there are }2\text{-}(v,k,1)\text{ designs for infinitely many }v.}
$$
The qualification at $k=1$ is necessary for the standard pair-design definition.
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