= Solution
Under the path convention used here, a real <Lévy process> $(L_a)_{a\geq0}$ starts at zero, has <independent increments> and <stationary increments>, is <stochastically continuous>, and has <càdlàg> paths almost surely. We verify all four requirements for $S$.
The immediate positive excursions proved in part (b) give $S_0=0$. For a deterministic $a\geq0$, $S_a$ is an almost-surely finite <stopping time>, and continuity gives $X_{S_a}=a$. The <Strong Markov property> implies that $\widetilde X_t=X_{S_a+t}-a$ is a fresh Brownian motion independent of the information up to $S_a$. For $b\geq0$,
$$
S_{a+b}-S_a=\inf\{t\geq0:\widetilde X_t>b\}.
$$
No crossing of $a+b$ could have occurred before $S_a$. The increment thus has the same law as $S_b$ and is independent of the past. Apply this successively at a finite increasing sequence of deterministic levels to prove stationary independent increments.
For $h>0$ and $\varepsilon>0$, part (b) and reflection give
$$
\mathbb P(S_h>\varepsilon)=2\Phi(h/\sqrt\varepsilon)-1\longrightarrow0
\quad(h\downarrow0).
$$
Hence $S_h\to0$ in probability. Stationarity of the increments gives stochastic continuity at every level, from either side where defined.
For path regularity, on a single probability-one event the <Brownian running maximum> $M_t=\max_{s\leq t}X_s$ is continuous, nondecreasing and unbounded, and
$$
S_a=\inf\{t:M_t>a\}.
$$
This inverse is nondecreasing. If $a_j\downarrow a$, then $S_{a_j}\downarrow L\geq S_a$. For any $t>S_a$, some $u<t$ has $M_u>a$; for all sufficiently large $j$, $M_u>a_j$, so $S_{a_j}\leq u<t$. Thus $L=S_a$, proving right continuity at every level simultaneously. On each bounded level interval it is bounded by the finite value at a larger integer level. Monotonicity then gives finite left limits. Therefore
$$
\boxed{S\text{ is a càdlàg Lévy process, the Brownian first-passage subordinator}.}
$$
The <Brownian first-passage subordinator> is the level-indexed increasing process, rather than a process indexed by the original Brownian time.
For $H$, the inverse relations for a continuous running maximum are
$$
H_a=S_{a-}\quad(a>0),\qquad H_{a+}=S_a.
$$
For example, if $b<a$ then $S_b\leq H_a$; while for $t<H_a$, $M_t<a$ and any $b\in(M_t,a)$ has $S_b>t$. Taking limits gives $S_{a-}=H_a$. Similarly $H_b\geq S_a$ for $b>a$, and a crossing of any level strictly above $a$ before a given $t>S_a$ bounds $H_b$ from above for $b$ sufficiently close to $a$. This proves $H_{a+}=S_a$. These are the <non-strict inverse of a continuous nondecreasing function> identities.
At the random level $A$ in part (b), $H_A<S_A=H_{A+}$, so $H$ fails right continuity almost surely. Thus it is not a <Lévy process> under the stated càdlàg-path convention. It does have the same <finite-dimensional distributions> and stochastic continuity as $S$, so definitions of a Lévy process that require only the increment and stochastic-continuity properties would call $H$ a Lévy process with a càdlàg modification. The reason for excluding the particular version $H$ here is its path regularity.
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