= Solution
The <boundary> assertion is interpreted at positive finite times: a chordal <SLE> already starts on the <domain boundary>, and its marked target is another <boundary> point. The precise conclusion in the <complex upper half-plane> is $\gamma(0,\infty)\subseteq\mathbb H$ almost surely.
For $0<\kappa\leq4$ and a fixed real $x>0$, the centered <Boundary-point Bessel flow for SLE> $V_t=g_t(x)-U_t$ solves
$$
dV_t=\frac2{V_t}\,dt-\sqrt\kappa\,dB_t.
$$
Thus $R_t=V_t/\sqrt\kappa$ is a <Bessel process> driven by $-B$ of dimension
$$
\boxed{\delta=1+\frac4\kappa\geq2.}
$$
The <Hitting-zero classification for a Bessel process> says that, started positively, it never hits zero when $\delta\geq2$. For $x<0$, reflect the equation and apply the same result. Therefore every fixed nonzero <boundary> point has infinite swallowing time. A countable intersection gives this simultaneously for all nonzero rational <boundary> points.
To extend this countable conclusion to every real $x>0$, choose rational $q$ with $0<q<x$. The difference of two centered boundary solutions satisfies
$$
\frac{d}{dt}(V_t(x)-V_t(q))=-\frac{2(V_t(x)-V_t(q))}{V_t(x)V_t(q)},
$$
so it remains positive while both flows exist. On any finite time interval, $V_t(q)$ is bounded away from zero, hence so is $V_t(x)$. Moreover $V_t(x)=x-U_t+\int_0^t2/V_s(x)\,ds$ remains bounded there, so the differential equation continues for the whole interval. Its uniform separation from the <Loewner driving function> also allows each flow map to continue as a <holomorphic function> on a complex <neighborhood> of $x$. Thus $x$ is outside $K_t$ and cannot be visited by the <Loewner trace>. Reflection gives the negative real axis. This uses countably many <Bessel processes> followed by deterministic flow comparison, rather than an uncountable union of probability-zero events.
To exclude a return to the starting point, use the <boundary extension of the inverse map for a continuous Loewner trace>: $g_s^{-1}$ is continuous on the closed <complex upper half-plane> and maps $U_s$ to $\gamma(s)$. This follows from the continuous trace theorem and the <Caratheodory boundary extension theorem>; it does not assume simplicity. At each deterministic rational time $s$, the <Conformal Markov property of SLE> makes the mapped future a fresh <SLE>, so its trace lies in $\mathbb H\cup\{0\}$ by the preceding argument. Mapping back with $g_s^{-1}$ therefore shows
$$
\gamma[s,\infty)\cap(K_s\cup\mathbb R)\subseteq\{\gamma(s)\}.
$$
If a point $q=\gamma(r)$ were revisited at $t>r$, choose rational $s\in(r,t)$ with $\gamma(s)\ne q$. Such an $s$ exists because <continuity> and strictly increasing <half-plane capacity> forbid a constant trace on an interval. But $q\in K_s\cup\mathbb R$, contradicting the displayed inclusion. This excludes all self-contacts, including returns to zero, and proves the <simple curve> property needed when discussing later swallowing times.
For $\kappa=0$, the deterministic <Loewner trace> is $\gamma(t)=2i\sqrt t$, so the conclusion follows directly. Hence the <boundary-intersection threshold for SLE> gives
$$
\boxed{\gamma(t)\in\mathbb H\text{ for every }t>0\text{ almost surely, for }0\leq\kappa\leq4.}
$$
Mapping to another marked <simply connected domain> carries positive-time points into its <interior>. The literal statement including the starting point is false, since $\gamma(0)$ is prescribed to be on the <domain boundary>.
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