Sufficient condition for a simply connected subset of $\mathbb{R}^2$ to be contractible. Simply connected subsets of $\mathbb{R}^2$ are not necessarily contractible, as shown for example by the Warsaw/Polish circle (see the answer to this question).
However, the two seem intuitively quite close (in the plane). I'm wondering if there is some simple and not too strong additional condition we can impose so that we get contractibility. For example, does locally connected + simply connected imply contractible (for subsets of $\mathbb{R}^2$)? If not, maybe locally path-connected?
In case it can help you to determine what could qualify as a "simple" condition to me, I'm a computer science student with some basic understanding of (mostly point set) topology.
 A: A space is said to be weakly contractible if it is path-connected and has trivial homotopy groups in all dimensions. A consequence of Whitehead's theorem is that any weakly contractible CW complex is contractible.

A space which is homotopy equivalent to a CW complex is contractible if and only if it is weakly contractible.

Now, according to Milnor, a space has the homotopy type of a CW complex if and only if it has the homotopy type of an absolute neighbourhood retract (ANR). An ANR is a metric space which is a neighbourhood retract of any metric space containing it as a closed subset. In any case, by the above, the weakly contractible planar ANRs furnish a large class of interesting examples for you.
Now, criteria to easily identify ANR's is found in a theorem of Dugundji.

Theorem (Dugundji): An $n$-dimensional metric space is an ANR if and only if it is $LC^n$.

Here a space $X$ is $LC^n$ if whenever $x\in U\subseteq X$ with $U$ open, there is an open $V\subseteq U$ containing $x$ for which the inclusion induces trivial maps $\pi_k(V,x)\rightarrow \pi_k(U,x)$ for all $0\leq k\leq n$.
The last piece of the puzzle comes from Zastrow's theorem, as proved in this paper by Cannon, Connor and Zastrow.

Theorem (Zastrow): If $X\subseteq \mathbb{R}^2$ is any subset, then $\pi_k(X,x)=0$ for all $k\geq2$ and $x\in X$.

Putting the pieces together, we have for any $X\subseteq\mathbb{R}^2$ that $\dim X\leq 2$. Thus such an $X$ is an ANR if and only if it is $LC^2$. But according to Zastrow, $LC^2$ will follow immediately upon $X$ being $LC^1$. A convenient condition which enforces $LC^1$ is that $X$ be locally simply connected. This helps identify planar ANRs, and Zastrow identifies the acyclic ones.

A locally simply connected subspace of $\mathbb{R}^2$ is contractible if and only if it is simply connected.

As for your question as to whether simply connected and locally simply connected should be equivalent in planar sets to simply connected and locally path connected, I do not know.
