Can a connected planar compactum minus a point be totally disconnected? What the title said.  In a slightly more leisurely fashion:-

Let $X$ be a compact, connected subset of $\mathbb{R}^2$ with more than one point, and let $x\in X$.  Can $X\smallsetminus\{x\}$ be totally disconnected?

Note that the Knaster-Kuratowski fan shows that, in the absence of the compactness hypothesis, the answer can be 'yes'.
To give credit where it's due, this question was inspired by one that I was asked by Barry Simon.
 A: I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.
Proposition. Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.
Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.
Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step. 
Corollary. If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.
In particular, we cannot remove a closed totally disconnected set (e.g. a single point) from a continuum to make it totally disconnected.
A: Being planar has nothing to do with the problem. Suppose a totally disconnects $X$ and choose $b$ different from $a$.  By passing to a sub continuum, assume that no proper sub continuum contains both $a$ and $b$. Take non empty disjoint open sets $U$ and $V$ whose union is $ X\sim a$.  WLOG $b$ is in $U$, and observe that $U\cup \{a\}$ is closed and connected. 
A: Let denote by $U_n\subset \mathbb R^2$ a sequence of open bounded neigborhoods of $X$, so that 
$$U_{n+1}\subset U_n\ \ \text{and}\ \  \bigcap_n U_n=X.$$
We can assume that all $U_n$ are connceted and therefore path-connected. Coose a point $p\in X$ distict from $x$ and consider a sequence of paths $\gamma_n$ in $U_n$ from $p$ to $x$. Fix $\epsilon>0$ such that $\epsilon<|p-x|$.
For each path choose the smalest value $t_n\in[0,1]$ so that $|\gamma_n(t_n)-x|=\epsilon$.
The image $Z_n=\gamma([0,t_n])$ is connected compact set.
Let $Z$ be a Hausdorff limit of a subsequence of $Z_n$.
Note that $Z$ is a compact connected subset of $X$.
Clearly, $Z\not\ni x$ and it contains at least two points; a contradiction
A: Two great answers have already been given, and I don't claim to add much, but here is something anyway.
A totally disconnected locally compact Hausdorff space has a basis of clopen sets, according to Proposition 3.1.7 of Arhangel'skii and Tkachenko, for example.  A closed set in $X-\{a\}$ need not be closed in $X$, but if $X$ is a metric space then the clopen subsets of $X-\{a\}$ at positive distance to $a$ will be clopen in $X$.  Thus if $X$ is a compact metric space with more than one point and $X-\{a\}$ is totally disconnected, then $X$ is not connected.
