Compact and connected set in $\mathbb{R}^2$ which is locally path-connected at only one point? 
Given a compact and connected set $A$ in $\mathbb{R}^2$. Can it be locally path-connected at only one point? And can it be locally path-connected at every point except one?

My thought
First I thought of the topological sine curve, which is compact and connected but not locally path-connected. But it doesn't solve the problems. Then I tried to construct a space that is locally path-connected at every point except one as
$$
A=\{(t\cos x,t\sin x):\cos x\in \mathbb{Q} , t\in[0,1]\}
$$
But this is not compact.
Do such spaces exist? Any hints would be highly appreciated.
 A: freakish's example checks out, so credit to him. I'll just elaborate on his comment. 
Let $C \hookrightarrow \{0\} \times \mathbb{R} \subseteq \mathbb{R}^2$ the ternary cantor set. and let $\mathcal{C}(C)$ be the cone over it. That is 
$$\mathcal{C}(C) = (C \times I)/C \times \{1\} \subseteq \mathbb{R}^2~~.$$
Then $\mathcal{C}(C)$ is the quotient of a compact Hausdorff space by collapsing a closed subspace and thus also compact (and Hausdorff). The cone over any space is contractible (to the tip) and thus connected, but $\mathcal{C}(C)$ is locally path connected only at the tip $t := \pi( C \times \{1\})$ with $\pi$ being the projection $C \times I \rightarrow \mathcal{C}(C)$. 
Locally path connected at the tip: Let $U$ be a neighborhood of $t \in \mathcal{C}(C)$. Then since $\mathcal{C}(C)$ carries the subspace topology so there is some $\varepsilon > 0$ s.t. $B_{\varepsilon}(t) \cap \mathcal{C}(C) \subseteq U$. This neighborhood is contactible via the contraction on $\mathcal{C}(C)$ restricted to $B_{\varepsilon}(t) \cap \mathcal{C}(C) \subseteq U$.
Nowhere else locally path connected: Let $x = (x_1, x_2) \in C \times I$ with $x_1 \neq 1$ and let $\varepsilon > 0$ s.t. $1 \not\in (x_1 - \varepsilon, x_1 + \varepsilon)$. Then consider the open neighborhood $(x_1 - \varepsilon, x_1 + \varepsilon) \times C$ of $x$. Since $\{ 1 \} \times C \cap (x_1 - \varepsilon, x_1 + \varepsilon) \times C =  \emptyset$, applying $\pi$ to this set yields a homeomorphism of an open subset of $\mathcal{C}(C)$ and $(x_1 - \varepsilon, x_1 + \varepsilon) \times C$. The latter is not locally path connected since $C$ is nowhere locally path connected.
A: At least subspaces reduced to a single point satisfy the requirements of the question.
