Must every compact and connected metric space be locally connected at at least one of its points?

There is a "folklore" counterexample. Peter Nyikos gives the construction here (see the last paragraph for the compactness) 


One more picture: BrouwerJaniszewskiKnaster continuum: 


In the product of the closet unit interval $I$ with the Cantor set $C$, identify $(0,x)$ and $\big(1,f(x)\big)$ where $f(x):=3x \mod 1$. The resulting space $X$ is the mapping cylinder of the continuous map $f:C\to C$. It is a compact metric space locally homeomorphic to $]0,1[\times C\, ,$ thus not locally connected at any point. Endpoints of $I$ have not a special role; we may equivalently obtain $X$ with a larger quotient, $(\mathbb{R}\times C) / \{ (t,x)=\big (t+1,f(x)\big) \} $. The important feature of the map $f:C\to C$ is that it has a dense orbit $f^n( x_0)$. This is easily seen as it is conjugate to the leftshift map on binary strings on the space ${\bf 2}^\mathbb{N}$, $(c_1,c_2,\dots)\mapsto(c_2,c_3,\dots)$ which is just how we see $f$ on the 2digits representations of points of the Cantor set. As a consequence, the image of $\mathbb{R}\times \{x_0\}$ in the latter quotient is a pathconnected dense subset of $X$, which is therefore connected. edit. Actually, such spaces are quite common in dynamical systems; an other example is the SmaleWilliams Solenoid and several strange attractors. 


The two examples given so far do not contain an illustration of the set. So here is one drawn with a "Cantorpen": 


No, examples abound, e.g., the $\sin\frac1x$curve, i.e., the closure of $\lbrace \sin\frac1x : 0 < x \le 1\rbrace$ in the plane. As noted below, this is not a good example (I misread the question). However, every indecomposable continuum, such as Knaster's buckethandle continuum, is an example because every proper subcontinuum is nowhere dense. The pseudoarc is, of course the ultimate example. 

