Give an example of non-normal subspace of a normal space. We know that a closed subspace of normal space is normal.  
My question was: why should other subspaces not work and then I came up with a counterexample.
It is peculiar that any subspace of regular space is regular.

Since $\mathbf{R}$ is a locally compact Hausdorff space, its $\mathbf{1}-$ point compactification $\mathbf{R^{*}}=\mathbf{R} \cup \lbrace \infty \rbrace$ gives a compact Hausdorff space(and hence normal).
Let $\mathbf{A}=\lbrace \infty \rbrace \cup \mathbf{Q}$ be a subspace of $\mathbf{R^{*}}$ . $\lbrace \mathbf{\infty} \rbrace$ and $(-\infty,0] \cap \mathbf{Q} $ are disjoint closed sets in $\mathbf{A}$  since  $\lbrace \mathbf{\infty}\rbrace=\mathbf{R^\mathsf{c}}$ and  $(-\infty,0] \cap \mathbf{Q}=(-\infty,0] \cap ( \mathbf{Q}  \cup \lbrace \infty \rbrace ) = (-\infty,0] \cap \mathbf{A} $ are closed in $\mathbf{R^{*}}$.
If $\mathbf{A}$ was normal as a subspace then we can find disjoint open sets in $\mathbf{U_{1},U_{2}}$ in $\mathbf{A}$  such that $\lbrace \infty \rbrace \subset \mathbf{U_{1}}$ and  $(-\infty,0] \cap \mathbf{Q}\subset \mathbf{U_{2}}$.
Then $\mathbf{U_{1}} = \mathbf{G_{1}} \cap \mathbf{A} $ and $\mathbf{U_{2}} = \mathbf{G_{2}} \cap \mathbf{A} $ such that $\mathbf{G_{1}}$ and $\mathbf{G_{2}}$ are open in $\mathbf{R^{*}}$.Let $\mathbf{H_{1}}$ and $\mathbf{H_{2}}$ be the basic open sets contained in $\mathbf{G_{1}}$ and $\mathbf{G_{2}}$ respectively such that $\lbrace \infty \rbrace \subset \mathbf{H_{1}} \cap \mathbf{A} \subset \mathbf{G_{1}} \cap \mathbf{A} $ and  $(-\infty,0] \cap \mathbf{Q} \subset \mathbf{H_{2}} \cap \mathbf{A} \subset \mathbf{G_{2}} \cap \mathbf{A}$.   Since $\infty \in \mathbf{H_{1}} \Rightarrow \mathbf{H_{1}^\mathsf{c}}$ is compact in $\mathbf{R}$. Also $(-\infty,0] \cap \mathbf{Q}=\lbrace $all non-positive rationals$\rbrace \subset \mathbf{H_{2}} \Rightarrow \mathbf{H_{2}}$ is open in $\mathbf{R}$ or $\mathbf{H_{2}^\mathsf{c}}$ is compact in $\mathbf
{R}$.
Since $\mathbf{U_{1}} \cap \mathbf{U_{2}} =\phi \Rightarrow (\mathbf{H_{1}} \cap \mathbf{A}) \bigcap (\mathbf{H_{2}} \cap \mathbf{A}) = \phi  \Rightarrow \mathbf{H_{1}} \cap \mathbf{H_{2}} \cap \mathbf{A} = \phi \Rightarrow \mathbf{H_{1}^\mathsf{c}} \cup (\mathbf{H_{2}} \cap \mathbf{A})=\mathbf{R^{*}} \ $.
But $\mathbf{H_{1}^\mathsf{c}}$ is compact in $\mathbf{R}$ implies that it is closed and bounded. In particular $\mathbf{H_{1}^\mathsf{c}}=[a,b]$ for some $a,b\in \mathbf{R}$. But [a,b] $\cup   (\mathbf{H_{2}} \cap \mathbf{A}) \subset [a,b] \cap \mathbf{A} = [a,b] \cup \mathbf{Q} \cup \lbrace \infty \rbrace \ne \mathbf{R^{*}}$ sincew we can find an irrational in $\mathbf{R}$ such that it is not in [a,b]. I guess we are done.THIS IS PLAIN WRONG>THIS SPACE IS METRIZABLE> 
