Show that there exists a path of class $C^1$ between $x$ and $y$ 
Let $H$ be an arc connectedness (a subset of $R^n$) , and both $x$ and $y$ components of $H$
Show that there exists a path of class $C^1$ between $x$ and $y$

I feel it's not true,
if we take for $H$ the graph of a function nowhere differentiable and we have $x '(t)$ somewhere in the middle of your path $t \mapsto (x (t), y (t) )$, then it bugue violently we must have $x'(u) \geq \varepsilon> 0$ on any open interval, so $x$ locally admits a differentiable inverse function, then there is nowhere differentiable on this interval. So you're required to have $x=0$, and suddenly it bugue also because $x$ is necessarily constant.

If I add $H$ is open, does it work ?

I feel the answer is yes, but I cannot prove it.
Thanks
 A: When $X$ is an arbitrary topological space, there is no notion of a $C^1$-map 
to $H$. Thus, I assume that $H$ is a subset of $R^n$ for some $n$ with the subspace topology. For $H$ open in $R^n$, the Weierstrass approximation theorem shows that you can approximate any continuous path in $H$ by a smooth one (with the same end-points). If $H$ is not required to be open then the answer is negative as we can take $H$ to be a planar Jordan curve of positive Hausdorff measure locally everywhere (say, Koch snowflake). Every subarc of this curve will have infinite length. Thus, there are no smooth nonconstant functions from $[0,1]$ to such $H$. 
Edit. Suppose that $f: [0,1]\to R$ is  continuous function and $f_n: [0,1]\to R$ 
is a approximating sequence. I assume that $f(0)=a\ne b=f(1)$ and will leave you to work out the case of equality. For $a_n=f_n(0), b_n=f_n(1)$ consider affine  functions  $h_n: R\to R$ 
$$
h_n(a_n)=a, h_n(b_n)=b. 
$$
You can write down these functions explicitly and verify that they converge (uniformly on compacts) to the identity function. Therefore, the compositions 
$$
g_n=h_n\circ f_n
$$
have fixed values at $0$ and $1$ and converge uniformly to $f$. 
