I previously asked In which topological spaces does the existence of a loop not contractable to a point imply there is a non-contractable simple loop also?
Given the broad scope of this question I propose this special case as an independent question:
In a path-connected subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point?
One-dimensional metric spaces and planar sets do have the property that you're interested in. To explain why this works out in such generality requires a combination of planar topology, continuum theory, and shape theory.
One-Dimensional Case: This is pretty classical, going back to work of Curtis and Fort in the 1950's.
Proof Sketch. Suppose $X$ is a one-dimensional metric space. If $X$ contains a simple closed curve, then any loop parameterizing that curve cannot be null-homotopic. A rigorous argument for this requires the shape injectivity of one-dimensional spaces. On the other hand, if $X$ contains no simple closed curves, then the image of any loop in $X$ is a dendrite, which is contractible. Thus either $X$ contains a non-contractible simple closed curve or is simply connected.
Planar Case: All the ingredients to prove what you want (and more) can be found in
H. Fischer, A. Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655--1676.
Proof Sketch. Suppose $X\subseteq \mathbb{R}^2$ is path connected and $\alpha:S^1\to X $ is not null-homotopic. Let $Y=\alpha(S^1)\cup\bigcup\{U\mid U\text{ is a bounded path component of }\mathbb{R}^2\backslash \alpha(S^1)\text{ with }U\subseteq X\}$. Fischer and Zastrow show $Y$ is a Peano continuum homotopy equivalent to a “Sierpinski-like” space, i.e. a copy of the Sierpinski Carpet with some, possibly empty, set of (originally deleted) squares filled back in. Since all we want to do is find some simple closed curve that doesn’t contract, pick any bounded path component $C$ of $\mathbb{R}^2\backslash Y$ and a point $z\in C\backslash X$. While $C$ is homeomorphic to an open disk, its boundary need not be a simple closed curve. However, you basically have a planar Peano continuum $Y$ surrounding a “hole” at $z$. Since $Y\subseteq X$ and $z\notin X$, any simple closed curve in $Y$ with winding number $1$ around $z$ will do the trick. If you want to create one explicitly, use some planar geometry to cover $Y$ with enough arc-wise connected open (in $Y$) sets to create a finite loop of neighborhoods in $Y$ with winding number $1$ around $z$ and build your curve piece-wise. This simple closed curve can't contract in $X$ since then it would contract in $\mathbb{R}^2\backslash \{z\}$, which clearly is impossible.
