Dense countable set in an infinite Baire and $T_1$ space has dense complement I'm right now covering chapter 8 of Topology by James R. Munkres, 2nd edition, and am stuck with the following problem. 
If $X$ is an infinite, $T_1$ and Baire space with a subset $D$ dense and countable, then $D^c$ is dense and $D$ cannot be an intersection of countable open sets. 
My approach: 
Let $D = \{d_1,d_2,\dots\}$ If $X$ is T1, then every set $U_n = \{d_1,\dots,d_n\}$ is closed. 
$$D^c = \bigcap_{i=1}^\infty U_n^c$$
Every $U_n^c$ is dense and open, then by Baire property $D^c$ is dense. I don't know if this is right, specially because I'm not sure if the $U_n^c$ are dense. 
For the other part of the problem, the hint says we can take $D = \bigcap W_n$ with $W_n$ open sets and consider the sets $V_d = X-\{d\}$ and to prove that $V_d$ and $W_n$ are dense. I'm not sure why this is true or why this is sufficient to prove the statement.
Thanks in advance.  
 A: As has been noted in the comments, the result is false as stated. In this case one can discover exactly what is going on by trying to complete the proof suggested by the hint and seeing where it runs into a problem.
Let $X$ be a $T_1$ space with a countable dense subset $D$. Suppose that there is a countable family $\mathscr{W}$ of open sets such that $D=\bigcap\mathscr{W}$. For each $d\in D$ let $V_d=X\setminus\{d\}$; $X$ is $T_1$, so each singleton $\{d\}$ is closed, and therefore each $V_d$ is open. Let $\mathscr{V}=\{V_d:d\in D\}$; then $\mathscr{V}$ is a countable collection of open sets, and $\bigcap\mathscr{V}=X\setminus D$. Thus, $\mathscr{W}\cup\mathscr{V}$ is a countable collection of open sets whose intersection is empty: 
$$\bigcap(\mathscr{W}\cup\mathscr{V})=\bigcap\mathscr{W}\cap\bigcap\mathscr{V}=D\cap(X\setminus D)=\varnothing\;.$$
If $X$ is a Baire space, this implies that at least one member of $\mathscr{W}\cup\mathscr{V}$ must fail to be dense in $X$. Clearly each $W\in\mathscr{W}$ is dense in $X$: if $U$ is any non-empty open set, then 
$$U\cap W\supseteq U\cap D\ne\varnothing\;.$$
Thus, if $X$ is a Baire space there must be at least one $d_0\in D$ such that $V_{d_0}$ is not dense in $X$. Then there is a non-empty open $U$ such that 
$$\varnothing=U\cap V_{d_0}=U\cap(X\setminus\{d_0\})=U\setminus\{d_0\}\;,$$
so $U=\{d_0\}$, and $d_0$ is an isolated point of $X$.
If, on the other hand, $X$ has no isolated points, then each of the sets $V_d$ is dense in $X$, $\mathscr{W}\cup\mathscr{V}$ is a countable family of dense open sets whose intersection is empty, and $X$ is not a Baire space.
In short, the result in the exercise is true if and only if $X$ has no isolated points. This exception is not really surprising, since — as was also mentioned in the comments — a dense set in a space must contain every isolated point of that space.
