I thought a more complete treatment of these beautiful and perhaps surprising results (Thms. 2 and 3) might be welcome. I have not provided proofs of Prop. 1 - 3, as they are standard exercises. (I will be glad to supply proofs upon request.)
First, some definitions:
Suppose $X$ is a topological space and $E \subset X$.
$E^c := $the complement of $E$.
int$(E) := \overline {E^c}^c$.
bd$(E) := \overline E \cap \overline {E^c}$.
$E$ is dense if $\overline E = X$.
$E$ is nowhere dense (nwd) if int($\overline E$) is empty.
$E$ is meager (aka 'first category') if it is a countable union of nwd sets.
$X$ is Baire provided that if $\{O_i\}$ is any collection of dense open subsets of $X$, then $\cap O_i$ is dense.
Prop. 1. If $E$ is open or closed, then bd($E$) is nwd.
Prop. 2. The following are equivalent:
(a) $E$ is nwd.
(b) If $O$ is open and non-empty, there exists an open non-empty set $P \subset O$ disjoint from $E$.
(c) If $O$ is open and $O \cap E$ is non-empty, there exists an open non-empty set $P \subset O$ disjoint from $E$.
Prop. 3. Suppose $A \subset B \subset X$.
(a) If $B$ is meager in $X$, then $A$ is meager in $X$.
(b) If $A$ is meager in $B$, then $A$ is meager in $X$.
Prop. 4. The following are equivalent:
(a) $X$ is Baire.
(b) If $O$ is open and non-empty, then $O$ is not meager.
Proof: This is a standard result. For a fairly sketchy proof, see this article. $\Box$
Prop. 5. Any open subset of a Baire space is Baire in the induced topology.
Proof: Suppose $X$ is Baire, $O$ is open in $X, P$ is open in $O$, and $P$ is non-empty. Then $P$ is open in $X$, so by Prop. 4(b), $P$ is not meager in $X$. Therefore by Prop. 3(b), $P$ is not meager in $O$. Again by Prop. 4(b), $O$ is Baire. $\Box$
Thm. 1 (Banach Category Theorem). The union of any family of open meager subsets of $X$ is meager.
Proof: (I follow Oxtoby here, with some minor modifications and more details.)
Let $M_0$ be such a family. $M := \cup M_0.$ Suppose $G_0 = \{U_\alpha \ | \ \alpha \in A\}$ is a maximal family of disjoint open non-empty sets with the property that each is contained in some member of $M_0. \ G := \cup G_0.$
We claim that $H := M \cap G^c$ is nwd. Otherwise, there exists $O$ open and non-empty with $O \subset \overline H \subset \overline M \cap G^c. \ P := O \cap M \subset G^c$, so $P$ is open and disjoint from all members of $G_0$. Since $\emptyset \neq O \subset \overline M, P$ is a non-empty subset of $M$. Therefore, there exists $S \in M_0$ with $T := P \cap S \neq \emptyset$. Then $T \cup G_0$ has the required properties and is a larger family than $G_0$ (#), establishing our claim.
For any $U_\alpha \in G_0$, we note that $U_\alpha$, being a subset of some set in $M_0$ is meager by Prop. 3(a). Suppose $U_\alpha = \cup N_{\alpha i}$ is a countable union of nwd sets. $N_i := \cup_ {\alpha \in A} N_{\alpha i}.$
We claim that $N_i$ is nwd. Suppose $O$ is open and $O \cap N_i \neq \emptyset$. Then there exists $\alpha$ such that $O \cap N_{\alpha i} \neq \emptyset. \ P := O \cap U_\alpha \ $. Since $U_\alpha$ is open, so is $P$. Observe that $O \cap N_{\alpha i} \subset O \cap U_\alpha = P$, whence $P \neq \emptyset$. Since $N_{\alpha i}$ is nwd., by Prop. 2(b), there exists $Q$ open with $\emptyset \neq Q \subset P$ and $Q \cap N_{\alpha i} = \emptyset$. Since $Q \subset P \subset U_a$ and the $U_a$'s are disjoint, $Q \cap U_\beta = \emptyset$ for $\beta \neq \alpha$, whence $Q \cap N_{\beta i} = \emptyset \ $. It follows that $Q \cap N_i = \emptyset \ $. Clearly $Q \subset P \subset O$, so Prop. 2(c) is satisfied by $O, Q,$ and $N_i$, establishing our claim.
Since $G = \cup G_0 = \cup N_i$, it is meager. Now $G \subset M$, so $M = (M \cap G^c) \cup G$. Thus $M$, being the union of a nwd subset and a meager one, is also meager. $\Box$
Thm. 2. $X$ contains an open meager subset whose complement is Baire in the induced topology.
Proof: Let $M_0$ consist of all open meager subsets of X. $\ M := \cup M_0.$ By Thm. 1, $M$ is meager. We will show that $B := M^c$ is Baire in the topology induced on $B$ by $X$. By Prop. 4(b), to prove this it suffices to show that if $O$ is open and non-empty in $B$, then $O$ is not meager in $B$. Suppose such an $O$ is meager in $B$. By Prop. 3(b), $O$ is meager in $X$. Now $O = P \cap B$, where $P$ is open in $X. \ P \cap M$, being a subset of $M$, is meager in $X$ by Prop. 3(a). Therefore, $P = (P \cap B) \cup (P \cap M) = O \cup (P \cap M)$ is open and meager in X, whence $P \in M_0$. It follows that $P \subset M$, whence $O \subset M$. Since $O \subset B$, we must have $O = \emptyset$ (#). $\Box$
Thm. 3. $X$ contains a closed meager subset whose complement is Baire in the induced topology.
Proof: Let $M$ and $B$ be as in the proof of Thm. 2. $F := \overline M = M \cup \textrm {bd}(M). \ M$ is open and meager, and by Prop. 1, bd($M$) is nwd. Therefore, $F$ is meager. $O := F^c \subset M^c = B. \ O$ is open in $X$, therefore in $B$. In the proof of Thm. 2 we showed that $B$ is Baire. By Prop. 5, $O$ is Baire. $\Box$