= Solution
For $\kappa>0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt\kappa$. The <Chordal Loewner equation> and $dU_t=\sqrt\kappa\,dB_t$ give
$$
dX_t=\frac{2}{\kappa X_t}\,dt-dB_t.
$$
Since $-B$ is again a standard <Brownian motion>, this is the <Boundary-point Bessel flow for SLE> with
$$
\boxed{\delta=1+\frac4\kappa.}
$$
For $x<0$, it is $-X_t$ that is the nonnegative <Bessel process>, with the same dimension; the formula without this sign convention is a signed Bessel flow.
When $0<\kappa\leq4$, one has $\delta\geq2$, and the <Hitting-zero classification for a Bessel process> says that $X_t$ never reaches zero. The borderline $\delta=2$ uses the <scale function of a one-dimensional diffusion> $\log x$: for $0<r<x<R$,
$$
\mathbb P_x(\tau_r<\tau_R)=\frac{\log R-\log x}{\log R-\log r}\longrightarrow0\quad(r\downarrow0).
$$
Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.
Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb R\setminus\{0\}$.
The same argument can be applied to the future after every rational time, using the <Conformal Markov property of SLE>. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual <crosscut> argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times proves
$$
\boxed{\operatorname{SLE}_\kappa\text{ is simple for }0<\kappa\leq4.}
$$
For $\kappa=0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt t]$. The division by $\sqrt\kappa$ is unnecessary in that case.
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