What is the difference between a convex set and a simply connected set as a subset of topological space, and as a subset of $\mathbb{R}^n$? 
What is the difference between a convex set and a simply connected
  set, as a subset of topological space, and as a subset of
  $\mathbb{R}^n$ in the usual topology ?

I mean, as far as I understood, a convex connected set is simply connected and vice versa, but if this were to be true, why there are two notion with different names ?
Note that the emphasise of the question is on the fact that the set is simply connected. I mean of course a convex set is connected, but what about its simpleness ?
 A: Simply connected means something else: that any loops inside the set can be continuously deformed to a point (as well as the set bring connected in the usual sense). So, for instance, an annulus in $\Bbb R^2$ is not simply connected because a loop going around the ring can't be continuously deformed to a point.
A convex set in $\Bbb R^n$ must be simply connected, but a simply connected set needs not be convex. For instance, a star-shaped set in the plane is simply connected but not convex.
A: You're  wrong: a simply connected set is not necessarily convex. For instance, a star-domain (also called radially convex set) is simply connected, and not necessarily convex:

A: 
as a subset of topological space

First of all note that you need a notion of vector space over $\mathbb{R}$ in order to talk about convexity. Perhaps it can be generalized to some bigger class of spaces but definitely not to all topological spaces. Unlike the notion of being simply connected.

a convex connected set is simply connected and vice versa

Convex set is simply connected (even contractible). But what makes you think that the converse holds? For example let $V$ be a real normed vector space and for any $v,w\in V$ consider
$$[v,w]=\big\{x\in V\ |\ x=tv+(1-t)w\text{ for some } t\in[0,1]\big\}$$
i.e. the line segment between $v$ and $w$. With that the definition of convexity is quite simple: $A\subseteq V$ is convex iff $[v,w]\subseteq A$ for any $v,w\in A$.
Now take two vectors $v,w\in V$ such that $\{v, w\}$ is linearly independent (for that you need $\dim(V)>1$). Then
$$A=[v,0]\cup[0,w]$$
is simply connected (it is even contractible since $\{0\}$ is a deformation retract of $A$) but not convex since $[v,w]\not\subseteq A$ (because of the linear independence).
