does simply connectedness require connectedness? My question consists of two parts.
$1)$
suppose domain $D=\{(x,y)\in\mathbb R^2~|~xy>0\}$ is given. Now that is first quadrant and third quadrant with exclusion of $x$ and $y$ axis. We can easily see that $D$ is not connected, since there is a discontinuity at origin. But every closed curve we can construct in domain contains interior of it (or formally, they can shrunk to a point). So do we call it simply connected, or do we also need connectedness to say $D$ is simply connected?
$2)$
Now for second part, let origin also included in domain so that $D$ is connected. Let us construct closed curve which goes through origin. Now this curve also satisfies assumption given above. But it is not simple closed curve. Does this affects simply connectedness? Do we need simple closed curves for simply connectedness?

A domain $D$ is defined as an connected, open subset of a finite-dimensional vector space. So if you have a simple connected domain it must be connected; thus $D$ in your first question is not a domain. And the paths do not need to be simple closed.
But in your second question you still do not have a domain if you include the origin, as $$D=\{(x,y)\in\mathbb R^2~|~xy>0\}\cup\{(0,0)\}$$ is not open.
Edit: why is $$D=\{(x,y)\in\mathbb R^2~|~xy>0\}\cup\{(0,0)\}$$ not a domain?
By including the origin we now have a connected set, but it is not open. For every $\varepsilon>0$ we have that $$(0,-\varepsilon)\in B_{2\varepsilon}(0,0)\quad\text{but}\quad(0,-\varepsilon)\notin D.$$ This means that we can't find $\varepsilon>0$ such that $B_{\varepsilon}(0,0)$ is contained in $D$. Thus $D$ is not open.
