= Solution
Before $T(x)$, the centered image satisfies the <stochastic differential equation>
$$
dX_t=\frac2{X_t}\,dt-\sqrt\kappa\,d\beta_t,\qquad X_0=x>0.
$$
Changing the sign of the <Brownian motion> shows that $X/\sqrt\kappa$ is a <Bessel process> of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.
Put $p=1-4/\kappa>0$. By the <Itô formula>, the drift of $X^p$ is
$$
pX^{p-2}\left(2+\frac\kappa2(p-1)\right)=0.
$$
Thus $X^p$ is a <local martingale> and a <scale function of a one-dimensional diffusion>. Fix $R>x$. The nonnegative function
$$
u_R(v)=\frac{R^{2-p}v^p-v^2}{\kappa+4},\qquad 0\leq v\leq R,
$$
vanishes at both endpoints and satisfies $\mathcal Lu_R=-1$, for $\mathcal L=(\kappa/2)\partial_{vv}+(2/v)\partial_v$. Stop first on $[\epsilon,R]$, and also at a deterministic time. The <Itô formula> gives $\mathbb E(t\wedge\tau_{\epsilon,R})\leq u_R(x)$. Let $t\to\infty$ and then $\epsilon\downarrow0$. It follows that the first exit $\tau_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.
<Optional stopping> applied to $X^p$ on this bounded interval gives
$$
\mathbb P_x(\tau_R<\tau_0)=\left(\frac{x}{R}\right)^p.
$$
On $\{\tau_0=\infty\}$ the process must therefore hit every $R>x$ before zero. Letting $R\to\infty$ makes that probability zero. Consequently
$$
\boxed{\kappa>4\ \Longrightarrow\ T(x)<\infty\quad\text{almost surely}.}
$$
At $\kappa=4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.
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