I was trying to solve the following question :
Let $X\subset \Bbb R \times \Bbb R$ be the union of $\Bbb R \times 0$ and $0 \times \Bbb R$. Let $\pi : X \to \Bbb R$ be the projection map of the first coordinate restricted to $X$ . Prove that $\pi$ is a closed map but not an open map.
I believe I can do the problem but I am facing difficulty in understanding what the underlying topology is i.e. whether it is the subspace topology or the product topology or something like the disjoint union topology.
So if someone suggest me what precisely are the open sets in X, then I could try the problem myself.
EDIT :
After considering the topology suggested in the comments, I have tried the problem.
$A := \{(x,y)\in X: x=0, 1<y<2\}$ is an open set in $X$ since it can be obtained as (in particular) $\{(x,y)\in X: -1 <x<1, 1<y<2\} \cap X$ with the former being open in $\Bbb R^2$ . But, $\pi(A) = 0$ and thus $\pi(A)$ is closed in $\Bbb R$ !
To show that $\pi$ is a closed map, A closed set in $X$ is of the form $C^/ =C \cap X$ where $C$ is a closed set in $\Bbb R^2$ .Since a closed set in $\Bbb R^2$ is of the form $C_1 \times C_2$ where $C_1,C_2$ are closed in $\Bbb R$.
Thus $\pi(C^/)=\pi(C\cap X)=\pi((C_1 \times C_2)\cap X)= C_1$ which is closed in $\Bbb R$ , so it's a closed map.
Can someone please check the proof and please point out mistakes.
That's not what you were supposed to prove. You were supposed to show that $\pi(A)$ is not open. Note that being closed is not equivalent to not being open. And indeed in every space $Y$ at least $\emptyset$ and $Y$ are both open and closed. In connected spaces like $\mathbb{R}$ or your $X$ these are only possibilites so your argument implies that $\pi(A)$ is not open (because it is neither empty nor whole $X$). Note that you would need to prove that $X$ is connected (which is simple because there's an obvious path from anywhere to $(0,0)$).
This is wrong. Have a look at the closed ball $\{(x,y)\ |\ x^2+y^2\leq 1\}$. It is not of the form $C_1\times C_2$.
The proper characterization is that every closed subset of $A\times B$ is a complement of an open subset. And open susbsets of $A\times B$ are of the form $\bigcup_{i}U_i\times V_i$ where each $U_i\subseteq A$ and $V_i\subseteq B$ are open.
Also your argument is invalid because it doesn't consider $X$ at all. And note that not all projections are closed, e.g. for $X=\{(x,y)\ |\ xy=1\}$ the projection onto first coordinate is not closed. It maps closed subset $\{(x,y)\ |\ xy=1;\ x<0\}$ onto $(-\infty, 0)$ which is not closed.
Now back to your problem: let $\pi:X\to\mathbb{R}$ be given by $\pi(x,y)=x$. What you need to show is that $C\subseteq X$ is closed if and only if $C=C_1\times\{0\}\cup\{0\}\times C_2$ where each $C_1,C_2\subseteq\mathbb{R}$ are closed (possibly empty). This is only because $X$ is what it is, it won't work for other definitions of $X$.
If $C_2$ is empty, then $\pi(C_1\times\{0\})=C_1$ is obviously closed. On the other hand if $C_2$ is not empty then
$$\pi(C)=\pi(C_1\times\{0\}\cup\{0\}\times C_2)=\pi(C_1\times\{0\})\cup\{0\}=C_1\cup\{0\}$$
and thus it is closed in $\mathbb{R}$.