Why singleton set is open in discrete metric space and not open in usual metric space explain? Let $X=(\mathbb R,d)$ be the usual real line and $Y=(\mathbb R,d′)$ be the set $\mathbb R$ with discrete metric.
Show that identity map from $X$ to $Y$ is not continuous but open as well as closed.
On the other hand, the identity map from $Y$ to $X$ is continuous which is neither open nor closed.
My attempt : we know that for continuity we want to show that inverse image of the open set is open.
Any singleton set $\{x\}$ is open in discrete metric space and hence its inverse image under identity map is also $\{x\},$ which is not an open set in usual metric.
Hence, identity map is not continuous.
Again if identity map have domain with discrete metric then it is always continuous.
Am I correct??
Now for open map (closed), we want to show that image of open (closed) set is open (closed).
How to use this definition to prove open and closed. Please help. Thank you.
 A: You have the right ideas.

Any singleton set $\{x\}$ is open in discrete metric space and hence its inverse image under identity map is also $\{x\}$, which is not an open set in usual metric.
Hence, identity map is not continuous.

Correct.

Again if identity map have domain with discrete metric then it is always continuous.

This is also correct, because any set in the discrete metric is open (and closed). More generally, let $f$ be a function from a discrete metric space $(M,\delta)$ to any metric space $(N,\delta')$, then for any open set $S\subseteq N$ we have $f^{-1}(S)\subseteq M$ open because $M$ is discrete, hence $f$ is continuous.

Now for open map (closed), we want to show that image of open (closed) set is open (closed).

Well, the reasoning is the basically the same as the one we used right now, isn't it? Given a function $g$ from $(N,\delta')$ to discrete $(M,\delta)$, let $S\subseteq N$ be open (closed). Now, what can we say about $g(S)$ and why?
A: In this answer I focus on the question: which of the identity map is open/closed and which is not. It seems to me that your ideas about continuity are okay.

Let $U$ be an arbitrary open set in $X$. 
Then $U$ (i.e. its image under the indentity map) is open in $Y$ since every set is  open in discrete space $Y$.
Proved is now that the map $\mathsf{id}:X\to Y$ is open.
Conversely take some singleton $\{y\}\subseteq Y$.
This is an open set in $Y$ but $\{y\}$ (the image under the identity map) is not open in $X$ (The title of your question suggests that you want a proof of that. Is that so?).
Proved is now that the map $\mathsf{id}:Y\to X$ is not open.

Let $U$ be an arbitrary closed set in $X$. 
Then $U$ (i.e. its image under the indentity map) is closed in $Y$ since every set is  closed in discrete space $Y$.
Proved is now that the map $\mathsf{id}:X\to Y$ is closed.
Conversely let $A$ be a set that is not closed in $X$.
This is a closed set in $Y$ since every set in $Y$ is closed. However, $A$ is not closed in $X$.
Proved is now that the map $\mathsf{id}:Y\to X$ is not closed.
