Always check the definitions being used in the reference - there are even significant differences between different arXiv versions of HAG2. Once you know you have an epimorphism from a union of affines, saying that $X$ is $n$-geometric in the HAG2 v7 sense basically amounts to saying that the higher diagonal
$$
X \to \mathrm{map}(S^{n},X)
$$
is affine. Thus $0$-geometric is equivalent to semi-separated, and any algebraic space $X$ is $1$-geometric because $\mathrm{map}(S^1,X)\cong X$.

EDIT: in response to keaton's comment, the condition isn't quite equivalent to $n$-geometricity, as there are some epimorphism conditions to check, but arises inductively because
$$
X\times^h_{\mathrm{map}(S^{n},X)}X \cong \mathrm{map}(S^{n+1},X).
$$

FURTHER EDIT with more details now I have time:

If you take affines $U,V$ etale over $X$, you want to show that $U\times_XV$ is $0$-geometric, and you already know that it is a scheme. Since the map $U\times_XV \to U \times V$ is a pullback of the diagonal $X \to X \times X$ of $X$, the relative diagonal $U\times_XV \to (U\times_XV)\times^h_{U \times V}(U\times_XV)$ is a pullback of $X \to \mathrm{map}(S^1,X)$, so is an isomorphism. The (absolute) diagonal of $U\times_XV$ is then a pullback of the diagonal of $U\times V$, hence affine.