Let ${\mathbf{x}}(s)$ be a curve in ${\mathbb{R}}^3$ with natural parameter $s$. We will need the following lemma, the proof of which is given at the end of this answer.
Lemma: Choose a fixed point ${\mathbf{y}}$. Then the curvature $\kappa$ satisfies
$$
\kappa \ge \left|\dot{\theta} + \frac{1}{r}\sin{\theta}\right|,
$$
where $r = \lVert\mathbf{x} - \mathbf{y}\rVert$ and $\theta$ is the angle between the
velocity $\dot{\mathbf{x}}$ and $\mathbf{x} - \mathbf{y}$.
Take $\mathbf{y} = \mathbf{x}(0)$; then $r(0)=0$ and $\theta(0)=0$. The function $r(s)$ is monotonically increasing as long as $0 \le \theta < \pi/2$ (since $\dot{r}(s) = \cos{\theta}$), so $\theta$ and $\kappa$ can be considered as single-valued functions of $r$ until that point. The above lemma then gives
$$
\kappa(r) \ge \left|{\theta}'(r)\dot{r} + \frac{1}{r}\sin{\theta(r)}\right| = \left|{\theta}'(r)\cos{\theta(r)} + \frac{1}{r}\sin{\theta(r)}\right| = \left|\frac{1}{r}(r \sin\theta(r) )'\right|.
$$
Until the first turning point of the motion (where the velocity becomes perpendicular to the radius), we have
$$
R\sin\theta(R) \le \int_{0}^{R} r \kappa(r) dr.
$$
If the curvature is strictly below a fixed value (say, $\kappa < K$), then the integral is less than $\frac{1}{2}KR^2$ for $R>0$, and we have the result that
$$
\sin{\theta(r)} < \frac{1}{2}Kr
$$
for $r>0$. A turning point is reached when $\theta=\pi/2$; this equation shows that the first such turning point must be at a radius greater than $2/K$, and hence the curve cannot be confined within a ball of diameter $2/K$. Finally, the arclength before reaching a given radius $R$ is bounded by
$$
\begin{eqnarray}
s(R) &=& \int_{0}^{R} \frac{ds}{dr}dr \\ &=& \int_{0}^{R} \frac{dr}{\cos\theta(r)} \\ &<& \int_{0}^{R} \frac{dr}{\sqrt{1 - \frac{1}{4}K^2 r^2}} \\ &=& \frac{2}{K}\sin^{-1}\left(\frac{1}{2}KR\right)
\end{eqnarray}
$$
for $R \le 2/K$. We conclude that any curve contained in the open unit ball with curvature $\kappa < 1$ must have length $s(2) < 2\sin^{-1}(1) = \pi$. Moreover, this bound is tight, since a circular arc joining the points at $\pm (1-\epsilon^2)\hat{\mathbf{z}}$ and the point at $(1-\epsilon)\hat{\mathbf{x}}$ has length approaching $\pi$ as $\epsilon \rightarrow 0$.
Proof of Lemma:
We will work in spherical coordinates centered at $\mathbf{y}$; then
$$
\begin{eqnarray}
{\mathbf{x}}
&=& r\hat{\mathbf{r}}, \\
{\dot{\mathbf{x}}}
&=& \dot{r}{\hat{\mathbf{r}}} + r\dot{\hat{\mathbf{r}}} \\
&=& \dot{r}{\hat{\mathbf{r}}} + r v_{\perp} \hat{\mathbf{v}}_{\perp}.
\end{eqnarray}
$$
Here $\hat{\mathbf{r}}$ is the unit vector from the origin to ${\mathbf{x}}$, and $\dot{\hat{\mathbf{r}}} = v_{\perp} \hat{\mathbf{v}}_{\perp}$ is its rate of change. Because $\hat{\mathbf{r}}$ has constant length, we have $\hat{\mathbf{v}}_{\perp}\cdot \hat{\mathbf{r}} = 0$. Taking the time derivative of this gives
$$
0 = \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}} + \hat{\mathbf{v}}_{\perp}\cdot \dot{\hat{\mathbf{r}}} = v_{\perp} + \dot{\hat{\mathbf{v}}}_{\perp}\cdot \hat{\mathbf{r}},
$$
which we will use later. Now, because $s$ is a natural parameter, $$\lVert\dot{\mathbf{x}}\rVert^2 = \left(\dot{r}\right)^2 + \left(rv_{\perp}\right)^2 = 1;$$
so we can define $\theta \in [0,\pi]$ such that $\dot{r} = \cos{\theta}$ and $rv_{\perp} = \sin{\theta}$. The velocity and acceleration become
$$
\begin{eqnarray}
{\dot{\mathbf{x}}}
&=& \left(\cos{\theta}\right){\hat{\mathbf{r}}} + \left(\sin{\theta}\right)\hat{\mathbf{v}}_{\perp}, \\
{\ddot{\mathbf{x}}}
&=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\dot{\hat{\mathbf{r}}} + \left(\cos{\theta}\dot{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp} \\
&=& -\left(\sin{\theta}\dot{\theta}\right){\hat{\mathbf{r}}} + \left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)\hat{\mathbf{v}}_{\perp} + \left(\sin{\theta}\right)\dot{\hat{\mathbf{v}}}_{\perp},
\end{eqnarray}
$$
and the acceleration has (two of its three) components
$$
\begin{eqnarray}
{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}} &=& -\left(\sin{\theta}\dot{\theta}\right) + \left(\sin{\theta}\right)\left(\dot{\hat{\mathbf{v}}}_{\perp} \cdot \hat{\mathbf{r}}\right) \\
&=& -\left(\sin{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right), \\
{\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp} &=& +\left(\cos{\theta}\right)\left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right).
\end{eqnarray}
$$
This brings us to the result that the squared curvature
$$
\kappa^2 = \lVert\ddot{\mathbf{x}}\rVert^2 \ge \left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{r}}\right)^{2} +
\left({\ddot{\mathbf{x}}}\cdot\hat{\mathbf{v}}_{\perp}\right)^{2} = \left(\dot{\theta} + \frac{1}{r}\sin{\theta}\right)^{2},
$$
where $\theta$ is the angle between the velocity and the outward radial vector. The lemma follows by taking the square root of both sides.