= Solution
A standard <Brownian motion> is a real <stochastic process> $(B_t)_{t\geq0}$ with $B_0=0$ <almost surely>, continuous <sample paths> <almost surely>, and independent increments such that $B_t-B_s$ has <normal distribution> $N(0,t-s)$ whenever $0\leq s<t$. With its completed, right-continuous <natural filtration> $(\mathcal F_t)$, its <Strong Markov property> is the following: for every almost-surely finite <stopping time> $T$, the process
$$
W_s=B_{T+s}-B_T,\qquad s\geq0,
$$
is a standard <Brownian motion> independent of $\mathcal F_T$. Here the <stopping-time sigma-algebra> is
$$
\mathcal F_T=\{A\in\mathcal F:A\cap\{T\leq t\}\in\mathcal F_t\text{ for all }t\geq0\}.
$$
In particular, for every bounded $\mathcal F_T$-measurable <random variable> $H$ and bounded Borel functional $G$ on the continuous path space,
$$
\mathbb E\bigl[H\,G((B_{T+s}-B_T)_{s\geq0})\bigr]
=\mathbb E[H]\,\mathbb E\bigl[G((B_s)_{s\geq0})\bigr].
$$
Thus the independence is from the entire stopped past, not only from $B_T$, and is a statement about the whole future path, not just a single increment.
To prove the <Brownian reflection principle> without assuming that a level is hit in finite time, fix $t>0$, $m>0$, and set $T_m=\inf\{s\geq0:B_s\geq m\}$ and $R=T_m\wedge t$. Continuity of the <sample paths> makes $T_m$ a <stopping time> and gives $B_{T_m}=m$ whenever $T_m<\infty$. The <stopping time> $R$ is bounded. Reflect the path after $R$:
$$
\widehat B_s=
\begin{cases}
B_s,&s\leq R,\\
2B_R-B_s,&s>R.
\end{cases}
$$
By the <Strong Markov property>, the future increments after $R$ are an independent <Brownian motion>. Negating them preserves their <probability distribution>, by symmetry of their joint <normal distributions>. Hence the reflected process $\widehat B$ has the same path law as $B$.
If $T_m>t$, reflection changes nothing on $[0,t]$ and $\widehat B_t=B_t<m$. If $T_m\leq t$, both paths first hit $m$ at the same time and $\widehat B_t=2m-B_t$. Therefore, for $x\leq m$, the following <events> agree path by path:
$$
\{M_t\geq m,\ B_t\leq x\}
=\{\widehat B_t\geq2m-x\}.
$$
Indeed, the right-hand endpoint is at least $m$, so the reflected path must have hit $m$ before or at $t$; this also forces the original path to have done so. The reflection relation then makes the endpoint inequality equivalent to $B_t\leq x$. Equality cases at $T_m=t$ satisfy the same relation. Since $\widehat B_t$ and $B_t$ have the same <probability distribution>,
$$
\boxed{\mathbb P(M_t\geq m,\ B_t\leq x)
=\mathbb P(B_t\geq2m-x)\qquad(m>0,\ x\leq m).}
$$
For $t=0$, both sides are zero, because $B_0=M_0=0$ and $2m-x\geq m>0$.
For $t>0$, set $x=m$ in this identity. Continuity implies $\{B_t>m\}\subseteq\{M_t\geq m\}$, and the <normal distribution> of $B_t$ has no atom at $m$. Splitting according to whether $B_t\leq m$ gives
$$
\mathbb P(M_t\geq m)=\mathbb P(B_t\geq m)+\mathbb P(B_t>m)
=2\mathbb P(B_t\geq m)=\mathbb P(|B_t|\geq m).
$$
Both <random variables> are nonnegative. Their tails agree for all $m>0$, which determines their <probability distributions>, including the mass at zero. Thus the <Brownian running maximum> satisfies
$$
\boxed{M_t\overset d=|B_t|\quad(t\geq0).}
$$
For $x>0$ and $t>0$, continuity again gives $\{T_x\leq t\}=\{M_t\geq x\}$. Since $B_t/\sqrt t$ is a <standard normal random variable>, the <distribution function> of the <Brownian first-passage time> is
$$
F_{T_x}(t)=2\left[1-\Phi\!\left(\frac{x}{\sqrt t}\right)\right].
$$
It tends to one as $t\to\infty$ and to zero as $t\downarrow0$, proving almost-sure finiteness and absence of an atom at zero. For the <standard normal random variable> $B_1$,
$$
\mathbb P\!\left(\left(\frac{x}{B_1}\right)^2\leq t\right)
=\mathbb P\!\left(|B_1|\geq\frac{x}{\sqrt t}\right)
=2\left[1-\Phi\!\left(\frac{x}{\sqrt t}\right)\right].
$$
The value of this ratio on the zero-probability <event> $\{B_1=0\}$ may be defined arbitrarily. Matching the <distribution functions> proves
$$
\boxed{T_x\overset d=\left(\frac{x}{B_1}\right)^2.}
$$
Finally, differentiate the <distribution function> on $t>0$, using $\Phi'=\phi$ and $\frac{d}{dt}(x/\sqrt t)=-x/(2t^{3/2})$. The <first-passage-time density> is
$$
\boxed{f_{T_x}(t)=\frac{x}{\sqrt{2\pi}\,t^{3/2}}\exp\!\left(-\frac{x^2}{2t}\right)\quad(t>0),}
$$
and is zero for $t\leq0$. Its integral is one by the endpoint limits of $F_{T_x}$, so no missing mass at an infinite <hitting time> is present.
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