By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part} there is an open subspace $U \subset X$ corresponding to the points of $|X|$ where $f$ is quasi-finite. We have to prove \begin{enumerate} \item[(a)] the image of $|U| \to |Y'|$ is $|U'|$ for some open subspace $U'$ of $Y'$, \item[(b)] $U = f^{-1}(U')$, and \item[(c)] $U \to U'$ is an isomorphism. \end{enumerate} Since formation of $U$ commutes with arbitrary base change (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part}), since formation of the normalization $Y'$ commutes with smooth base change (Lemma \ref{lemma-normalization-smooth-localization}), since \'etale morphisms are open, and since ``being an isomorphism'' is fpqc local on the base (Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-isomorphism}), it suffices to prove (a), (b), (c) \'etale locally on $Y$ (some details omitted). Thus we may assume $Y$ is an affine scheme. This implies that $Y'$ is an (affine) scheme as well. \medskip\noindent Let $x \in |U|$. Claim: there exists an open neighbourhood $f'(x) \in V \subset Y'$ such that $(f')^{-1}V \to V$ is an isomorphism. We first prove the claim implies the lemma. Namely, then $(f')^{-1}V \cong V$ is a scheme (as an open of $Y'$), locally of finite type over $Y$ (as an open subspace of $X$), and for $v \in V$ the residue field extension $\kappa(v) \supset \kappa(\nu(v))$ is algebraic (as $V \subset Y'$ and $Y'$ is integral over $Y$). Hence the fibres of $V \to Y$ are discrete (Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}) and $(f')^{-1}V \to Y$ is locally quasi-finite (Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-fibres}). This implies $(f')^{-1}V \subset U$ and $V \subset U'$. Since $x$ was arbitrary we see that (a), (b), and (c) are true. \medskip\noindent Let $y = f(x) \in |Y|$. Let $(T, t) \to (Y, y)$ be an \'etale morphism of pointed schemes. Denote by a subscript ${}_T$ the base change to $T$. Let $z \in X_T$ be a point in the fibre $X_t$ lying over $x$. Note that $U_T \subset X_T$ is the set of points where $f_T$ is quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part}. Note that $$ X_T \xrightarrow{f'_T} Y'_T \xrightarrow{\nu_T} T $$ is the normalization of $T$ in $X_T$, see Lemma \ref{lemma-normalization-smooth-localization}. Suppose that the claim holds for $z \in U_T \subset X_T \to Y'_T \to T$, i.e., suppose that we can find an open neighbourhood $f'_T(z) \in V' \subset Y'_T$ such that $(f'_T)^{-1}V' \to V'$ is an isomorphism. The morphism $Y'_T \to Y'$ is \'etale hence the image $V \subset Y'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_T(z) \in V'$. Observe that $$ \xymatrix{ (f'_T)^{-1}V' \ar[r] \ar[d] & (f')^{-1}(V) \ar[d] \\ V' \ar[r] & V } $$ is a fibre square (as $Y'_T \times_{Y'} X = X_T$). Since the left vertical arrow is an isomorphism and $\{V' \to V\}$ is a \'etale covering, we conclude that the right vertical arrow is an isomorphism by Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-isomorphism}. In other words, the claim holds for $x \in U \subset X \to Y' \to Y$. \medskip\noindent By the result of the previous paragraph to prove the claim for $x \in |U|$, we may replace $Y$ by an \'etale neighbourhood $T$ of $y = f(x)$ and $x$ by any point lying over $x$ in $T \times_Y X$. Thus we may assume there is a decomposition $$ X = V \amalg W $$ into open and closed subspaces where $V \to Y$ is finite and $x \in V$, see Lemma \ref{lemma-etale-splits-off-quasi-finite-part}. Since $X$ is a disjoint union of $V$ and $W$ over $Y$ and since $V \to Y$ is finite we see that the normalization of $Y$ in $X$ is the morphism $$ X = V \amalg W \longrightarrow V \amalg W' \longrightarrow S $$ where $W'$ is the normalization of $Y$ in $W$, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-normalization-in-disjoint-union}, \ref{spaces-morphisms-lemma-finite-integral}, and \ref{spaces-morphisms-lemma-normalization-in-integral}. The claim follows and we win.