We need a Lemma first (which is exercise $ 2.5.2 $ in Qing Liu's book).
Lemma. Let $ X $ be a scheme and $ x \in X $. Then, $ \text{codim} ( \overline { \left \{ x \right \} } , X ) = \dim \mathcal{O}_{X,x} $. If $ Z $ is a closed subset of $ X $, then $ \text{ codim} ( Z, X ) = \min _ { z \in Z } \dim \mathcal{O}_{X,z} $.
Proof. If $ \left \{ x \right \} \subseteq Z_{0} \subsetneq Z_{1} \ldots, \subsetneq Z_{n} = X $ is a sequence of irreducible closed subsets of $ X $, then, for any open $ U \subset X $ containing $ x $, we have $ \left \{ x \right \} \subseteq U \cap Z_{0} \subsetneq U \cap Z_{1} \subsetneq \ldots \subsetneq U \cap Z_{n} = U $. We may thus assume that $ U $ is the spectra of a ring $ A $, and $ x = \mathfrak{p} $ is a prime. Then, $ \overline { \left \{ x \right \} } = V ( \mathfrak{p} ) $ and $ \text { codim } ( V( \mathfrak{p} ) , X ) = \text{ht} ( \mathfrak{p} ) = \dim A_{ \mathfrak{p} } = \dim \mathcal{O} _ { X , x } $.
Let $ Z $ be a closed subset of $ X $. Let $ C_{i} $ be the irreducible closed subsets of $ Z $. Then, by definition, $ \text{codim} ( Z, X ) = \inf _{ i } \text{ codim } ( C_{i} , X ) $. It therefore suffices to assume that $ Z $ is irreducible. Moreover, if $ U_{j} $ is an open covering of $ Z $, then $ \text{codim} ( Z, X ) = \inf _{ j } \text{codim } ( Z \cap U _{ j } , U _{ j } ) $ for similar reasons as above. We may thus also assume that $ X $ is affine and $ Z $ is an irreducible closed subset i.e. $ X = $ Spec $ A $ and $ Z = V ( \mathfrak { p } ) $ for some prime $ \mathfrak { p } $ of $ A $. Then, $ \text {codim} ( Z, X ) = \text{ht} ( \mathfrak{p} ) = \dim A_ { \mathfrak { p } } = \min _ { \mathfrak{q} \supset \mathfrak{p} } \dim A_{\mathfrak{q} } = \min _ { z \in Z } \dim \mathcal{ O } _ { X , z } $. $ \quad \quad \quad \quad $ $ \square $
We now turn to the proof of the proposition.
$ \boxed { 3) \Longleftrightarrow 2) } $. By the Lemma, we see that $ \text{codim} ( X - U , x ) \geq 1 $ iff $ U $ contains points all the points of codimension $ 0 $, which are the same as generic points.
$ \boxed { 2) \implies 1 ) } $ If $ U $ contains all generic points, then any closed subset of $ X $ that contains $ U $ must contain all the generic points, and thus, must contain all the irreducible components of $ X $, and thus must equal $ X $.
Suppose now that $ X $ is locally Noetherian.
$ \boxed { 1) \implies 2) } $ Suppose that $ \eta $ is a generic point of $ X $ outside $ U $. Then, $ U \cap \overline { \left \{ \eta \right \} } = \emptyset $. Consider the set $ S = \left \{ Z | Z \text{ irreducible component of } X \right \} \setminus \overline { \left \{ \eta \right \} } $. Let $$ C = \bigcup _ { Z \in S } Z . $$
We claim that $ C $ is closed. This will result in a contradiction since $ C \supset U $ and $ C \neq X $. Let $ V $ be an open affine subset of $ X $. Then, since $ V $ is Noehterian, the union $ \bigcup _ { Z \in S } V \cap Z $ is a finite union of closed subsets of $ V $, and is thus closed in $ V $. Thus, $ C $ is closed in $ X $.
