Is a continuous image of a normal space normal? 
Problem Let $f:X\rightarrow Y$ be closed continuous surjective map. Show that if $X$ is normal then So is $Y$.

What if we drop the 'closed' condition? I want a counter example. I know the proof of this theorem.
 A: Let $X$ be $\Bbb N$ with the discrete topology, let $Y$ be $\Bbb N$ with the cofinite topology, and let $f:X\to Y$ be the identity map. Clearly $X$ is normal and $f$ is continuous and surjective. However, $Y$ is $T_1$ but not Hausdorff, so it clearly is not normal.
A: Consider $f\colon \Bbb R\amalg\Bbb R\to \Bbb R\amalg\Bbb R/\sim$, where $(x,0)\sim(x,1)$ for all $x\neq 0$. That is, $f$ is the quotient map to the line with a doubled-origin. Clearly $f$ is continuous and surjective (as are all quotient maps), but $f$ is not closed: take an interval $[0,1]$ in the first copy of $\Bbb R$. Then $f([0,1])$ is the interval $(0,1]$ plus one of the two origin points. This set is not closed since the preimage consists of $[0,1]$ in the first copy of $\Bbb R$ and $(0,1]$ in the second copy.
A: Let $Y$ be any topological space and let $X$ be $Y$ with the discrete topology.  Then $X$ is normal, and the identity map $X\to Y$ is a continuous surjection.  So, any non-normal space $Y$ at all will give a counterexample in this way.
(In fact, more strongly, every topological space is a quotient of a normal space.  See for instance the construction in this answer of mine.)
