In a topological space if there exists a loop that cannot be contracted to a point does there exist a simple loop that cannot be contracted also? I'm interested in whether one only needs to consider simple loops when proving results about simply connected spaces. 
If it is true that: 
In a Topological Space, if there exists a loop that cannot be contracted to a point then there exists a simple loop that cannot also be contracted to a point.
then we can replace a loop by a simple loop in the definition of simply connected.
If this theorem is not true for all spaces, then perhaps it is true for Hausdorff spaces or metric spaces or a subset of $\mathbb{R}^n$?
I have thought about the simplest non-trivial case which I believe would be a subset of $\mathbb{R}^2$. 
In this case I have a  quite elementary way to approach this which is to see that you can contract a loop by shrinking its simple loops. 
Take any loop, a continuous map, $f$,  from $[0,1]$. Go round the loop from 0 until you find a self intersection at $x \in (0,1]$ say, with the previous loop arc, $f([0,x])$ at a point $f(y)$ where $0<y<x$. Then $L=f([y,x])$ is a simple loop. Contract $L$ to a point and then apply the same process to $(x,1]$, iterating until we reach $f(1)$. At each stage we contract a simple loop. Eventually after a countably infinite number of contractions we have contracted the entire loop. We can construct a single homotopy out of these homotopies by making them maps on $[1/2^i,1/2^{i+1}]$ consecutively which allows one to fit them all into the unit interval.
So if you can't contract a given non-simple loop to a point but can contract any simple loop we have a contradiction which I think proves my claim. 
I'm not sure whether this same argument applied to more general spaces or whether it is in fact correct at all. I realise that non-simple loops can be phenomenally complex with highly non-smooth, fractal structure but I can't see an obvious reason why you can't do what I propose above. 
Update: Just added another question related to this about classifying the spaces where this might hold - In which topological spaces does the existence of a loop not contractible to a point imply there is a non-contractible simple loop also? 
 A: Here is an example of topological space $X$, embeddable as compact subspace of $\mathbf{R}^3$, that is not simply connected, but in which every simple loop is homotopic to a constant loop.
Namely, start from the Hawaiian earring $H$, with its singular point $w$. Let $C$ be the cone over $H$, namely $C=H\times [0,1]/H\times\{0\}$. Let $w$ be the image of $(w,1)$ in $C$. Finally, $X$ is the bouquet of two copies of $(C,w)$; this is a path-connected, locally path-connected, compact space, embeddable into $\mathbf{R}^3$.

It is classical that $X$ is not simply connected: this is an example of failure of a too naive version of van Kampen's theorem.
However every simple loop in $X$ is homotopic to a constant loop. Indeed, since the joining point $w\in X$ separates $X\smallsetminus\{w\}$ into two components, such a loop cannot pass through $w$ and hence is included in one of these two components, hence one of the two copies of the cone $C$, in which it can clearly be homotoped to the sharp point of the cone.
A: Every finite simplicial complex is weakly homotopy equivalent to a finite space. Therefore there are finite spaces with nontrivial loops; and these are obviously not embedded. 
A: This question came up when I was taking a course in topology in a bygone century. For homework I constructed an example of a subspace $X$ of $\mathbb R^3$ which is not simply connected although every simple closed curve in $X$ is homotopic to a point. It was something like this:
Take an infinite sequence of circles in the $xy$-plane, each circle externally tangent to the next one, with the centers of the circles lying on a straight line and converging to the origin. For concreteness, we may suppose that the $n^\text{th}$ circle is a circle of radius $\frac1{2^n}$ centered at $\left(\frac3{2^n},0\right)$. Make each of those circles the base of a right circular cone of height $1$. Finally, let $X$ be the closure of the union of that sequence of cones. Every simple closed curve in $X$ can be shrunk to a point in $X$, since it lies on one cone; but a closed curve which goes around the bases of all the cones cannot be shrunk to a point in $X$.

From the same course I vaguely recall a proposition to the effect that, if $X$ is "locally simply connected in the large" (meaning that each point has a neighborhood $U$ such that every closed curve in $U$ is homotopic to a point in $X$), and if every simple closed curve in $X$ is homotopic to a point, then $X$ is simply connected. I don't recall if there were other conditions on $X$ (such as "Hausdorff space" or "metric space"), and I certainly don't recall anything about the proof, except that it could not have been anything deep.
