Searching for a connected neighborhood of fixed radius for every point of a submanifold Let $\Gamma=\gamma(\partial D)\subset \mathbb R^3$ be an image of $C^1$-curve, where $D$ is an open ball in $\mathbb R^2$.
My question is whether it is possible to find a fixed $\epsilon >0$ such that for every point $p \in \Gamma$ the neighborhood $$B_\epsilon(p) \cap \Gamma$$ is connected in space $\Gamma$. $B_\epsilon(p)$ denotes open ball of radius $\epsilon$ in $\mathbb R^3$.
I thought since $\Gamma$ can be considered as $C^1$-manifold of $\mathbb R^3$ some neighborhood of every point $p$ can be approximated (i.e. diffeomorphic) to tangent line. Now repeat procedure for every point, take smallest and we get desired $\epsilon$.
But how to assure no pathologies like $\sin(1/x)$ can occure? In this case I would get $\epsilon = 0$.
Edit: As @MoisheCohen  pointed out we have to assume that 


*

*$\Gamma$ is a compact 1 dimensional $C^1$-submanifold of $\mathbb R^3$ i.e. exists $\gamma: S^1 \to \mathbb R^3 $ regular smooth injective map such that $\gamma(S^1)=\Gamma$.

 A: You should first prove that for any point $p \in \Gamma$ there exists an $\bar{\epsilon} = \bar{\epsilon}(p) > 0$ (which depends on $p$) such that $B_{\epsilon}(p) \cap \Gamma$ is connected for every $0 < \epsilon \le \bar{\epsilon}$.
Now observe that $\{B_{\epsilon}(p) \cap \Gamma  \colon  0 < \epsilon < \bar{\epsilon}(p)\}$ is an open cover of $\Gamma$. Then the conclusion follows from the compactness of $\Gamma$. 
A: Assume that $B(x, \epsilon)\bigcap \Gamma$ is not
connected for $x\in \Gamma$.
Hence there is at least four points in $\partial B(x,\epsilon
)\bigcap \Gamma $, i.e. $x_i$ where $x_1,\ x_2$ are connected in the
ball and so are $x_3,\ x_4$.
Here there is $x_{12}$ in the arc between $x_1$ and $x_2$ whose
tangent line $T_{12}$ is close to a line containing segment
$[x_1x_2]$.
Consider a sequence of points $a_n\in \Gamma$ converging to $a$ s.t.
$B(a_n,1/n)\bigcap \Gamma$ is not connected. 
Here $T_{12}(a_n),\ T_{34}(a_n)$ have different limits so that it is
not $C^1$.
If they have same limits and $x_2,\ x_3$ are connected, then there
is a point $x_{23}$ whose tangent line $T_{23}(a_n)$ is close to a
line containing $[x_2x_3]$. Hence $T_{12}(a_n),\ T_{23}(a_n)$ have
different limits. 
