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As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is one in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$ (i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.

EDIT: Thanks to Nate Eldredge for helping me to simplify my original argument.

As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is a normal space in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$ (i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.

EDIT: Thanks to Nate Eldredge for helping me to simplify my original argument.

As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is one in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$ (i.e., the set of all vertical lines contained in $U$) contains a dense open subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$.) By the Baire Category Theorem, any countable intersection of open neighborhoods of $C$ contains a horizontal line.

As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is one in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$ (i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.

EDIT: Thanks to Nate Eldredge for helping me to simplify my original argument.

As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is one in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = \{0,1\} \times [0,1]$ works. ($C$ is the two vertical lines on the far right and far left.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ [0,1] \times \{x\} \subseteq U\}$$ (i.e., the set of all horizontal lines contained in $U$) contains a dense open subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(0,p)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$.) By the Baire Category Theorem, any countable intersection of open neighborhoods of $C$ contains a horizontal line.

As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.

A perfectly normal space is one in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then $$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$ (i.e., the set of all vertical lines contained in $U$) contains a dense open subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$.) By the Baire Category Theorem, any countable intersection of open neighborhoods of $C$ contains a horizontal line.