Connected, locally connected, path-connected but not locally path-connected subspace of the plane I am looking for a set in the plane (with respect to the natural Euclidean topology) that is connected, locally connected, path-connected but not locally path-connected. I did not find one in Steen-Seebach, but maybe I overlooked?  

Here's an example in $\mathbb{R}^3$. The construction does not adapt to the plane, and I don't know if there do exist examples in $\mathbb{R}^2$.
Start by letting $A\subseteq\mathbb{R}^2$ be connected, locally connected and not locally path connected. For example, see my answer on mathoverflow for an example where $A$ is totally path disconnected (i.e., every continuous $f\colon\mathbb{R}\to A$ is constant), so has no path connected subsets other than single points.
Then,
$$
X = \left\{(x,y,z)\in\mathbb{R}^3\colon (y,z)\in A{\rm\ or\ }x=0\right\}
$$
is connected, locally connected, path connected but not locally path connected.
To see this, set
\begin{align}
Y &= \mathbb{R}\times A = \left\{(x,y,z)\in\mathbb{R}^3\colon (y,z)\in A\right\},\\
Z&=\{0\}\times\mathbb{R}^2=\left\{(x,y,z)\in\mathbb{R}^3\colon x=0\right\}
\end{align}
so that $X=Y\cup Z$. Then, $Y$ is locally connected, as $\mathbb{R}$ and $A$ satisfy these properties. So, $X$ is locally connected away from $Z$. Furthermore, we have a deformation retraction from $X$ onto $Z$ by $F((x,y,z),t)=(tx,y,z)$, so $X$ is contractible and, hence, path connected. Restricting to a neighbourhood of any point in $Z$, this also shows that $X$ is locally contractible at all points of $Z$, so $X$ is locally connected. However, if $A$ is not locally path connected at a point $p\in A$ then $X$ cannot be path connected about any point $(x,p)\in X$ for all $x\not=0$.
