Title basically says everything. Prove that if $u\in C^{2}(\mathbb{R}^{n}_{+})\cap C(\bar{\mathbb{R}^{n}_{+}})$ is a bounded solution of the BVP
$$\left\{\begin{array}
-\Delta u=0&\text{in}\;\mathbb{R}^{n}_{+}\\
u=g&\;\text{on}\;\partial\mathbb{R}^{n}_{+},
\end{array}\right.$$
then it is unique.
Various tools I have in mind are maximum principle, mean value formulas, Liouville's theorem, "energy" functionals, and Harnack's inequality, uniqueness of Green's function, Hopf's lemma, etc….but in all my scratch work to prove the problem, I keep running into technical difficulties in working with the boundary at infinity.
I arrived at a proof by using a result from Evans exercise #2.5.10 (Schwarz reflection principle):
Proof. Consider the ball $B_{R}(0)$ and suppose $u_{R}$ is a bounded solution to the above problem, but posed on the domain $B_{R}^{+}(0):=\{x:x\in B_{R}(0), x_{n}>0\}.$ Let $v_{R}$ be another bounded solution and define $w_{R}:v_{R}-u_{R}.$ Then $w_{R}=0$ on $\partial B^{+}_{R}(0)\cap\{x_{n}=0\}$, and the Schwarz reflection principle states that the odd extension of $w_{R}$ to $B^{-}_{R}(0)$ is harmonic in all of $B_{R}(0)$ (the proof of this is trivial if we assume $w\in C^{2}(\bar{B^{+}_{R}(0)})$ by using the mean-value formulas, and only a little more difficult under the present assumptions by using Poisson's formula for the ball). Now, $u_{R},v_{R}$ both being bounded implies $w_{R}$ is also bounded. Sending $R\to\infty$, we find that $w:=\lim_{R\to\infty}w_{R}$ is a bounded and harmonic in $\mathbb{R}^{n}$, from which it follows $w_{R}\equiv0$ by Liouville's theorem and the fact that $w=0$ on the hyperplane $x_{n}=0.$
Let $u_{1}$ be a bounded solution to the problem above. Suppose $u_{2}$ is another solution and define $$w:=u_{1}-u_{2}.$$ Then $w=0$ on $\partial\mathbb{R}^{n}_{+}.$
Best Answer
Added: For the half space $H$ there is a nice direct argument, using the following result from Chapter 2 of Harmonic Function Theory by Sheldon Axler, Paul Bourdon, and Wade Ramey.
Corollary 2.2. Suppose that $u$ is a continuous bounded function on $\bar H$ that is harmonic on $H$. If $u=0$ on $\partial H$, then $u\equiv 0$ on $\bar H$.
Proof: For $x\in \mathbb{R}^{n-1}$ and $y<0$ define $u(x,y)=-u(x,-y)$, thereby extending $u$ to a bounded continuous function defined on all of $\mathbb{R}^n$. Clearly $u$ satisfies the local mean-value property specified in Theorem 1.24, so $u$ is harmonic on $\mathbb{R}^n$. Liouville's Theorem (2.1) now shows that $u$ is constant on $\mathbb{R}^n$.
The uniqueness can be proved using Brownian motion $(B_t)$. Let $D$ be your domain and define $T_D=\inf(t>0: B_t\notin D)$ to be the first time when Brownian motion exits the domain. For $x\in D$, the process $u(B_{t\wedge T_D})$ is a bounded $\mathbb{P}_x$-martingale, so $u(x)=\mathbb{E}_x(u(B_{t\wedge T_D}))$. Since $u$ is bounded and $\mathbb{P}_x(T_D<\infty)$, we have no difficulty in letting $t\to\infty$ and concluding that $$u(x)=\mathbb{E}_x(g(B_{T_D})).$$
A similar argument is given in Proposition 2.6 in Chapter 4 of Brownian motion and classical potential theory by Sidney C. Port and Charles J. Stone. There they approximate $D$ from within by relatively compact regular open sets $D_n$, use uniqueness on each $D_n$, and take limits to obtain uniqueness on $D$.