Let C be a connected and completely metrizable subset of the Euclidean plane. Can C fail to be locally connected at each of its points?

The pseudoarc is another planar example. In fact, isn't Victor's example a pseudoarc? 


Edit Here is an easier example. Let $K$ be the middle third Cantor set and $$B=\{(t,ty): 0\leq t\leq 1, y\in K\}$$ be the cone over $K,$ the Cantor branch and $C=B\cup\varphi(B),$ where $\phi$ is the central symmetry about the midpoint $(1/2,0)$ of the bottom twig $[0,1]\times\{0\}.$ Thus $C$ is obtained by gluing two Cantor branches rotated by $\pi$ relative to each other along their bottom twigs. This space $C$ is compact and pathwise connected, but it is not locally connected at each point. Indeed, a small neighborhood of every point $P$ will contain parts of Cantor twigs not passing through $P.$ Yes. A solenoid is a homogeneous continuum (=compact connected metric space) embeddable in $\mathbb{R}^3$ that is not locally connected at any point, in fact, a small neighborhood of each point looks like the Cantor set crossed with an interval. Its generic projection to $\mathbb{R}^2$ is compact, connected, and not locally connected at each of its points. [Edit I am not longer confident that the last claim is true.] 


Take the subset $A=\left(\{1\}\times[0,1]\right)\cup\left(\bigcup_{n\in\mathbb{N}}[0,1]\times\{\frac{1}{n}\}\right)\cup\left([0,1]\times \{0\}\right)$. In this way describe, it might seem awful, but if you draw it, it's a very simple set. This set is closed, and so completely metrizable. It is also connected (three segments are enough to connect any two points, so it is even pathwise connected). But at the point $(0,0)$ it has no neighbourhood base of connected subsets. 


Any indecomposable continuum has the property you desire. (A continuum is indecomposable if it cannot be written as a union of two proper subcontinua.) One way to see this is that any indecomposable continuum has uncountably many composants, all of which are mutually disjoint, and all of which are dense in $x$. (Here the composant of a point $x$ in $X$ is the union of all proper subcontinua of $X$ that contain $x$.) Here is a more direct proof: Suppose that $C$ is a proper subcontinuum of $X$ that is a neighborhood of $x$. Then every connected component of $V := X\setminus C$ contains a point of $C$ in its boundary. (This is known as the 'boundary bumping theorem'.) If $\overline{V}$ is connected, then $\overline{V}$ and $C$ are proper subcontinua of $X$ whose union is $X$. If $V$ is disconnected, decompose $V$ into two relatively closed disjoint subsets $A$ and $B$; then $A\cup C$ and $B\cup C$ are the desired subcontinua. A simple example of an indecomposable continuum is given by the Knaster buckethandle, see http://commons.wikimedia.org/wiki/File%3aThe_Knaster_%22buckethandle%22_continuum.svg . The solenoid, mentioned in another answer, is another indecomposable continuum. You can also get such examples from "Lakes of Wada" continua. Of course the double Cantor brush given by Victor is not indecomposable (and in fact hereditarily decomposable). 

