The implication (2) $\Rightarrow$ (1) is Lemma \ref{lemma-properties-covering-imply-diagonal}. Assume (1). Let $x \in |\mathcal{X}|$ be a finite type point. We will produce a scheme over $\mathcal{X}$ which ``works'' in a neighbourhood of $x$. At the end of the proof we will take the disjoint union of all of these to conclude. \medskip\noindent Let $U$ be an affine scheme, $U \to \mathcal{X}$ a smooth morphism, and $u \in U$ a closed point which maps to $x$, see Lemma \ref{lemma-point-finite-type}. Denote $u = \Spec(\kappa(u))$ as usual. Consider the following commutative diagram $$ \xymatrix{ u \ar[d] & R \ar[l] \ar[d] \\ U \ar[d] & F \ar[d] \ar[l]^p \\ \mathcal{X} & u \ar[l] } $$ with both squares fibre product squares, in particular $R = u \times_\mathcal{X} u$. In the proof of Lemma \ref{lemma-point-finite-type-monomorphism} we have seen that $(u, R, s, t, c)$ is a groupoid in algebraic spaces with $s, t$ locally of finite type. Let $G \to u$ be the stabilizer group algebraic space (see Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-stabilizer-groupoid}). Note that $$ G = R \times_{(u \times u)} u = (u \times_\mathcal{X} u) \times_{(u \times u)} u = \mathcal{X} \times_{\mathcal{X} \times \mathcal{X}} u. $$ As $\mathcal{X}$ is quasi-DM we see that $G$ is locally quasi-finite over $u$. By More on Groupoids in Spaces, Lemma \ref{spaces-more-groupoids-lemma-groupoid-on-field-dimension-equal-stabilizer} we have $\dim(R) = 0$. \medskip\noindent Let $e : u \to R$ be the identity of the groupoid. Thus both compositions $u \to R \to u$ are equal to the identity morphism of $u$. Note that $R \subset F$ is a closed subspace as $u \subset U$ is a closed subscheme. Hence we can also think of $e$ as a point of $F$. Consider the maps of \'etale local rings $$ \mathcal{O}_{U, u} \xrightarrow{p^\sharp} \mathcal{O}_{F, \overline{e}} \longrightarrow \mathcal{O}_{R, \overline{e}} $$ Note that $\mathcal{O}_{R, \overline{e}}$ has dimension $0$ by the result of the first paragraph. On the other hand, the kernel of the second arrow is $p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{e}}$ as $R$ is cut out in $F$ by $\mathfrak m_u$. Thus we see that $$ \mathfrak m_{\overline{z}} = \sqrt{p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{e}}} $$ On the other hand, as the morphism $U \to \mathcal{X}$ is smooth we see that $F \to u$ is a smooth morphism of algebraic spaces. This means that $F$ is a regular algebraic space (Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}). Hence $\mathcal{O}_{F, \overline{e}}$ is a regular local ring (Properties of Spaces, Lemma \ref{spaces-properties-lemma-regular}). Note that a regular local ring is Cohen-Macaulay (Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}). Let $d = \dim(\mathcal{O}_{F, \overline{e}})$. By Algebra, Lemma \ref{algebra-lemma-find-sequence-image-regular} we can find $f_1, \ldots, f_d \in \mathcal{O}_{U, u}$ whose images $\varphi(f_1), \ldots, \varphi(f_d)$ form a regular sequence in $\mathcal{O}_{F, \overline{z}}$. By Lemma \ref{lemma-slice} after shrinking $U$ we may assume that $Z = V(f_1, \ldots, f_d) \to \mathcal{X}$ is flat and locally of finite presentation. Note that by construction $F_Z = Z \times_\mathcal{X} u$ is a closed subspace of $F = U \times_\mathcal{X} u$, that $e$ is a point of this closed subspace, and that $$ \dim(\mathcal{O}_{F_Z, \overline{e}}) = 0. $$ By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-fibre-at-a-point} it follows that $\dim_e(F_Z) = 0$ because the transcendence degree of $e$ relative to $u$ is zero. Hence it follows from Lemma \ref{lemma-quasi-finite-at-point} that after possibly shrinking $U$ the morphism $Z \to \mathcal{X}$ is locally quasi-finite. \medskip\noindent We conclude that for every finite type point $x$ of $\mathcal{X}$ there exists a locally quasi-finite, flat, locally finitely presented morphism $f_x : Z_x \to \mathcal{X}$ with $x$ in the image of $|f_x|$. Set $W = \coprod_x Z_x$ and $f = \coprod f_x$. Then $f$ is flat, locally of finite presentation, and locally quasi-finite. In particular the image of $|f|$ is open, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}. By construction the image contains all finite type points of $\mathcal{X}$, hence $f$ is surjective by Lemma \ref{lemma-enough-finite-type-points} (and Properties of Stacks, Lemma \ref{stacks-properties-lemma-characterize-surjective}).