By Mean Value property for harmonic functions,
$$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u dA$$
where $x \in R^2$ is a point, $B_r(x)$ is the ball of radius $r$ centred at $x$ and the integration is over area.
When $r \to \infty$, integral above becomes the average of $u$ over whole plain. Since this average is a constant, we conclude $u(x)$ should be a constant function.
To state this argument more rigorously; Let $x_1$ and $x_2$ be distinct in $R^2$ and $|x_1-x_2|=a$:
Then by Mean Value theorem
\begin{equation}u(x_2)=\frac{1}{\pi r^2}\int_{B_r(x_2)}u dA
\end{equation}
\begin{equation}u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA
\end{equation}
Where ${B_r(x_2)}$ is the $r$-ball around $x_2$ and ${B_{r+a}(x)}$ is the ball with radius $r+a$ around $x_1$.
Rewriting (2) we get:
$$u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA + \frac{1}{\pi (r+a)^2}\int_{S}u dA$$
Where $S=B_{r+a}(x_1)$\ $ B_{r}(x_2) \in R^2$ is the hashed area in the graph.
We shall prove $u(x_2)-u(x_1)=0$. Since (1) and (2) are true for all $r$, it suffices to show $$lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA- \frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA$$
$$=lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA-\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA - \frac{1}{\pi (r+a)^2}\int_{S}u dA=0$$
But first two terms cancel out in limit and remains
$$u(x_2)-u(x_1)=lim_{r \to \infty} \frac{-1}{\pi (r+a)^2}\int_{S}u dA=0$$
Since $u$ is bounded we can write
\begin{equation}
\begin{aligned}
lim_{r \to \infty} \frac{1}{\pi (r+a)^2}\int_{S}u dA & \leq lim_{r \to \infty} \frac{1}{\pi (r+a)^2} u_{max} S_a \\
& =lim_{r \to \infty} \frac{u_{max}[\pi (r+a)^2-\pi (r)^2]}{\pi (r+a)^2} \\
& =lim_{r \to \infty} \frac{u_{max}(a^2-2ar)}{(r+a)^2}=0 \\
\end{aligned}
\end{equation}
Where $u_{max}$ is the upper bound of $u$ and $S_a=\pi (r+a)^2-\pi (r)^2$ is the area of $S$.
Hence $u(x_2)-u(x_1)=0 \qquad \forall x_1,x_2 \in R^2$ and $u$ is constant.
