Let $U$ be an open, connected set in $\mathbb{R}^n$ and let $(B(t))_{t \geq 0}$ be an $n$-dimensional Brownian motion with start at $x \in U$ and let $\overline{B_x(\delta)}$ be the closed ball about $x$ of radius $\delta$ that is contained in $U$. In order to get from $x$ to the boundary $\partial U$, $B(t)$ must first pass through $\partial B_x(\delta)$. I'd like to know how this can be proved rigorously. This is the gist of the problem.
More exactly, following is a theorem from chapter 3, "Harmonic functions, transience and recurrence" of Mörters and Peres's textbook Brownian Motion (to be referred to below as [M]). The theorem is listed below together with the proof given in the textbook (but not verbatim). I'm interested in a rigorous justification for the last couple of equations in the end of the proof, namely $$ \begin{aligned} E_x\left[E_x\left[\varphi(B(\tau))\mathbb{1}_{\{\tau < \infty\}} \mid \mathcal{F}^+(\rho)\right]\right] & = E_x\left[u(B(\rho))\right] \\ & = \int_{\partial B_x(\delta)} u(y) \mu_{x,\delta}(dy) \end{aligned} $$
Setting the stage
Let $n \in \mathbb{N}_1$. Denote by $\mathbf{C}$ the set consisting of all continuous functions from $[0,\infty)$ to $\mathbb{R}^n$. For every $t \in [0,\infty)$, denote by $\pi_t : \mathbf{C} \rightarrow \mathbb{R}^n$ the projection on the $t$th coordinate and denote by $\mathcal{B}$ the minimal $\sigma$-algebra on $\mathbf{C}$ in which all the $\pi_t$'s are measurable. For every $x \in \mathbb{R}^n$ denote by $P_x$ the probability measure over the measurable space $(\Omega := \mathbf{C}, \mathcal{F} := \mathcal{B})$ that renders the stochastic process $(\pi_t)_{t \geq 0}$ a Brownian motion with start at $x$. For every $x \in \mathbb{R}^n$ and every $\mathcal{B}/\textrm{Borel}$-measurable function $\varphi: \mathbf{C} \rightarrow \mathbb{R}$, denote by $E_x(\varphi)$ the expectation of $\varphi$ with respect to the probability space $(\Omega, \mathcal{F}, P_x)$, provided this expectation exists (we allow the possibility that $E_x(\varphi) \in \{\pm \infty\}$).
The Theorem
([M] Theorem 3.8, p. 68)
Suppose $U$ is a non-empty, open, connected set in $\mathbb{R}^n$ and define $\tau$ to be the first hitting time of $U$'s boundary, i.e. $$ \tau := \inf \{t \geq 0 \mid : B(t) \in \partial U\} $$
Let $\varphi: \partial U \rightarrow \mathbb{R}$ be measurable w.r.t. the $\sigma$-algebra induced on $\partial U$ by $\textrm{Borel}(\mathbb{R}^n)$, and suppose the function $u: U \rightarrow \mathbb{R}$, defined by $$ u(x) := E_x[\varphi(B(\tau)) \mathbb{1}_{\{\tau < \infty\}}] $$ for every $x \in U$, is locally bounded, i.e. for every $x \in U$ there is some neighborhood $N$ of $x$ such that $N \subseteq U$ and $u$ is bounded on $N$. In particular, we suppose that the expectation on the right is well-defined for every $x \in \mathbb{R}^n$.
Then $u$ is a harmonic function.
Proof
([M] Proof of theorem 3.8, p. 68)
We start with a lemma.
Lemma ([M] Theorem 3.2, p. 65) Let $D \subseteq \mathbb{R}^n$ be a non-empty, open and connected set and let $v : D \rightarrow \mathbb{R}$ be a locally bounded function. Then $v$ is harmonic on $D$ iff $v$ has the (spherical) mean value property: For all $z \in D$ and $s \in (0,\infty)$ such that the closed ball $\overline{B_z(s)} \subseteq D$, we have: $v$ is integrable on $\partial B_z(s)$ and $$ v(z) = \int_{\partial B_z(s)} v(w) \mu_{z, s}(dw) $$ where $\mu_{z, s}$ is the normalized uniform measure on the sphere $\partial B_z(s)$.
Let $x \in \mathbb{R}^n$ and let $\delta \in (0,\infty)$ be such that the closed ball $\overline{B_x(r)} := \{y \in \mathbb{R}^n \mid: |y - x| \leq r\}$ is contained in $U$. Define the stopping time $\rho := \inf \{t \in (0,\infty) \mid: B(t) \neq B_x(\delta)\}$. Then the strong Markov property implies that $$ \begin{aligned} u(x) & = E_x\left[E_x\left[\varphi(B(\tau))\mathbb{1}_{\{\tau < \infty\}} \mid \mathcal{F}^+(\rho)\right]\right] \\ & = E_x\left[u(B(\rho))\right] \\ & = \int_{\partial B_x(\delta)} u(y) \mu_{x,\delta}(dy) \end{aligned} $$ where $\mu_{x,\delta}$ is the normalized uniform measure on the sphere $\partial B_x(\delta)$. Therefore, by the lemma, $u$ is harmonic. Q.E.D.
Taken for granted that $\rho \leq \tau$, let us see why the derivation you gave makes sense. Recall the equation chain:
$$ \begin{aligned} u(x) & \stackrel{A}{=} E_x\left[E_x\left[\varphi(B(\tau))\mathbb{1}_{\{\tau < \infty\}} \mid \mathcal{F}^+(\rho)\right]\right] \\ & \stackrel{B}{=} E_x\left[u(B(\rho))\right] \\ & \stackrel{C}{=} \int_{\partial B_x(\delta)} u(y) \mu_{x,\delta}(dy) \, .\end{aligned} $$
$A$: this is the definition of $u$ and a property of conditional expectation.
$B$: this is the strong Markov property and the definition of $u$, a bit more explicit:
I take some notation from the book of Revuz and Yor (Sections I.3 and III.3). Define for $x \in E^{\mathbb{R}_+}$ and $ t\geq 0$ the translation operator $\theta_t: \, E^{\mathbb{R}_+} \to E^{\mathbb{R}_+}, \, x \mapsto \theta_t(x)$ with $(\theta_t(x))(s)= x(t+s)$. Define $\tau \circ \theta_t = \inf \{ s \geq 0 : \, B(s+t) \in \partial U\}$ and set $\theta_T(\omega) = \theta_t(\omega)$ if $T(\omega) = t$ for a stopping time $T$. For the stopping times $\rho$ and $\tau$ it is true that $$ \tau = \rho + \tau \circ \theta_\rho \, , $$ since $\rho \leq \tau$. \begin{align} E_x\left[\varphi(B(\tau))\mathbb{1}_{\{\tau < \infty\}} \mid \mathcal{F}^+(\rho)\right] & = E_x\left[\varphi(B(\rho + \tau \circ \theta_\rho))\mathbb{1}_{\{\rho + \tau \circ \theta_\rho < \infty\}} \mid \mathcal{F}^+(\rho)\right] \\ & = E_{B(\rho)} \left[ \varphi(B(\tau)) \mathbb{1}(\tau < \infty) \right] \\ & = u(B(\rho)) \, . \end{align}
$C$: The distribution of Browian motion started at $x \in \mathbb{R}^d$ hitting the sphere $\partial B_x(\delta) = \{y \in \mathbb{R}^d: \, |y| = 1\}$ is uniform: $P_x(B(\rho) \in dy) = \mu_{x,\delta}(dy)$.
Regarding $\rho \leq \tau$ (if $\rho = \inf \{ t\geq 0: \, B(t) \in \partial B_x(\delta)\}$): We take a single path $X \in \mathbf{C}$ with $X(0)=x \in \mathbb{R}^d$ and show $\rho \leq \tau$ pathwise. If $\tau = \infty$, there is nothing to show. So we can assume that $X(t) \in \partial U$ for some $\tau = t>0$.
Let $t' = \sup \{ s\geq 0:\, X(s) \in |x-X(s)|<\delta \}$ and $t'' = \inf \{s\geq t':\, |x-X(s)|>\delta\}$. Clearly, $t'\leq t''$. Since the modulus is continuous we know that $s \mapsto d(s) = |x-X(s)|$ is continuous. We have $d(t')\leq \delta$ since there is a sequence $s_n \to t'$ non-decreasing with $d(s_n)<\delta$. Similarly $d(t'')\geq \delta$. If $t'=t''$ we are done, otherwise note that for $r \in (t',t'')$ we can neither have $d(r)<\delta$, nor $d(r)>\delta$, also leading to $d(r) = \delta$. In both cases we conclude that there is a $r = \rho\leq t$ with $X(r) \in \partial B_x(\delta)$. But this says that for this particular path $\rho \leq \tau$.