Indeed it is a mistake. If a topological space has a nonempty open Baire subspace, it is of the second category in itself. But if it also has a nonempty meagre open subset, it is not a Baire space.
Thus spaces like $[0,1] \cup \mathbb{Q}$, or the topological sum of a Baire space and a non-Baire space are non-Baire spaces that are of the second category in itself. (See below)
A correct condition would be that every nonempty open subset of $X$ is of the second category (in itself or in $X$, for open subspaces this is equivalent).
Let $X$ a topological space, and $U \subset X$ open. If $N\subset X$ is nowhere dense in $X$, then $N\cap U$ is nowhere dense in $U$. For suppose $V := \operatorname{int}_U(\operatorname{cl}_U(N\cap U)) \neq \varnothing$. Since $U$ is open, $V$ is also open in $X$, and $V \subset \operatorname{cl}_U(N\cap U) = \operatorname{cl}_X(N) \cap U \subset \operatorname{cl}_X(N)$, so if $N\cap U$ is not nowhere dense in $U$, then $N$ is not nowhere dense in $X$. And if $N \subset U$ is nowhere dense in $U$, then $N$ is nowhere dense in $X$. For if $V := \operatorname{int}_X(\operatorname{cl}_X(N)) \neq \varnothing$, then $V\cap U \neq \varnothing$ (since $\operatorname{cl}_X(N) \subset \operatorname{cl}_X(U)$), and $V\cap U \subset \operatorname{cl}_U(N)$, so then $N$ is not nowhere dense in $U$.
From that it follows that every open subspace of a Baire space is itself a Baire space, and that every space that has a nonempty open Baire subspace is of the second category in itself.