Find continuous functions $f,g$ such that $g\circ f $ is closed and continuous but neither $g$ nor $f$ is closed map. 
Find continuous functions $f,g$ such that $g\circ f $ is closed and continuous but neither $g$ nor $f$ is closed map.
Find continuous functions $f,g$ such that $g\circ f $ is open and continuous but neither $g$ nor $f$ is open map.
Consider the spaces to be $\Bbb R$

I took $f(x)=$ \begin{cases} (x-1)  &-1\le x\le 1\\0 & x>1,x<-1\end{cases}
and
$g(x)=$ \begin{cases} (x-1)  &x>1,x<-1\\0 & -1\le x\le 1\end{cases}
Though $g\circ f=0$ which is closed and $f$ is not closed since it takes $(-0.5,0.5)$ to $0$ but here $g$ is closed
I want examples where $g,f$ are not closed both. How to make $g$ not closed?
How to answer the second question?
 A: 
Find continuous functions $f,g$ such that $g\circ f $ is closed and continuous but neither $g$ nor $f$ is closed map.

We shall following a hint by user87690. First we put $f(x)=e^x$. Since $f(\Bbb R)=(0,\infty)$, the map $f$ is not closed. Now let $g(x)$ equals $1$, if $x\ge -1$ and $-1/x$, otherwise. Since $g(\Bbb R)=(0, 1]$, the map $g$ is not closed. But $gf(x)=1$, so $gf(A)=\{1\}$ for each non-empty closed subset $A$ of $\Bbb R$.

Find continuous functions $f,g$ such that $g\circ f $ is open and continuous but neither $g$ nor $f$ is open map.

There are no such functions. Indeed, assume to the contrary that the function $gf:\Bbb R\to\Bbb R$ is open. First we show that the function $gf$ in injective. Suppose to the contrary that there exists $x,x’\in\Bbb R$ such that $x<x’$ and $gf(x)=gf(x’)=y$. Since $gf$ is a continuous map, an image $gf([x,x’])$ of the segment $[x,x’]$ is compact. Therefore there exist numbers $m=\min gf([x,x’])$ and $M=\max gf([x,x’])$. If $m<y$ then an image $gf((x,x’))$ of an open set $(x,x’)$ equals to $[m,M]$ or to $[m,M)$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $M>y$. If $m=M=y$ then $gf((x,x’))=\{y\}$, a contradiction again. Thus $gf(x)\ne gf(x’)$. Thefore the function $f:\Bbb R\to\Bbb R$ in injective too. By invariance of domain, $f$ is an open map.
