The general one-particle spin state for a spin 1/2 particle is $$|\psi\rangle = a\mid\uparrow\rangle + b\mid\downarrow\rangle$$
with $|a|^2 + |b|^2 = 1$. So let us try to anti-symmetrize two of these. \begin{align*}\operatorname{Alt}(|\psi\rangle_1 \otimes |\psi\rangle_2) = & (a_1 \mid\uparrow\rangle + b_1\mid\downarrow\rangle)\otimes (a_2 \mid\uparrow\rangle + b_2 \mid\downarrow\rangle) - (a_2 \mid\uparrow\rangle + b_2 \mid\downarrow\rangle)\otimes (a_1 \mid\uparrow\rangle + b_1 \mid\downarrow\rangle)\\
= & (a_1a_2 - a_1a_2) \mid\uparrow\rangle\mid\uparrow \rangle + (b_1b_2 - b_2b_1)\mid\downarrow\rangle\mid\downarrow \rangle \\ & + (a_1b_2 - a_2 b_1)\mid \uparrow\rangle\mid\downarrow \rangle
+ (b_1a_2 - b_2 a_1) \mid\downarrow\rangle\mid\uparrow \rangle \\
= & (a_1b_2 - a_2b_1) (\mid\uparrow \downarrow \rangle - \mid\downarrow\uparrow\rangle)
\end{align*}
so whatever one-particle state you start with you end with something proportional to $\mid\uparrow \downarrow \rangle - \mid\downarrow\uparrow\rangle$.
More abstractly, if $v_1, v_2, \ldots, v_n$ are a basis for a vector space $V$, a basis the anti-symmetric rank 2 tensors on $V$ is given by $$v_i \otimes v_j - v_j \otimes v_i \quad 1 \le i < j \le n.$$
In the case that $ n = 2$, this reduces to the previous result.
