A weird subset of $\mathbb R^2$

Is there a path-connected subset of $\mathbb R^2$ such that any path connecting 2 distinct points in that subset has infinite length? I am told that there is such a set, but I don't know what it is. Thank you.

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Something like a Koch snowflake should work. – Chris Eagle Feb 13 '12 at 20:33
There is, for example almost every path of a standard Brownian motion. – Did Feb 13 '12 at 20:34
Would $\{x \}$ count? – Matt Feb 13 '12 at 20:36
@Matt I don't believe that's in the spirit of the question. – Austin Mohr Feb 13 '12 at 20:46

The graph of any continuous nowhere differentiable function $\mathbb R\to\mathbb R$ is an example (or any continuous function that is not of bounded variation on any subinterval).

If $f:[a,b]\to\mathbb R$ is of bounded variation, then $f'$ exists in $(a,b)$ except on a set of measure $0$. Thus if $f:\mathbb R\to\mathbb R$ is nowhere differentiable, then $f$ is not of bounded variation on any subinterval.

Suppose that $f:\mathbb R\to\mathbb R$ is continuous and nowhere differentiable. Then $G=\{(x,f(x)):x\in\mathbb R\}\subset \mathbb R^2$ is path connected, and if $a<b$, then the length of any path in $G$ connecting $(a,f(a))$ to $(b,f(b))$ is bounded below by the total variation of $f$ on $[a,b]$, hence is infinite.

Exercise 26 in Chapter 3 on page 49 of Wheeden and Zygmund's Measure and integral gives a hint of a way to construct a continuous function on $[0,1]$ that is not of bounded variation on any subinterval, with no reference to differentiation, using a modification of the construction of the Cantor function.