MORE GENERAL AND SIMPLE ANSWER
Liouville Type (Theorem's) Suppose $u:\mathbb R^d\to \mathbb R$ be a nonegaitve harmonic function Then, $u$ is constant.
Proof $u\ge 0$ then $\inf_{\mathbb R^d} u<\infty$. Hence we set $v= u-\inf_{\mathbb R^d} u$ and
\begin{split} \begin{cases}\Delta v=0\\ v\ge 0\\ \inf_{\mathbb R^d} v =0\end{cases}\end{split}
Let $\varepsilon>0,$ and $y\in\mathbb R^d$ then there exists $x_\varepsilon$ such that,
$ v(x_\varepsilon)\le \inf_{\mathbb R^d} v+\varepsilon$
Let $R>\max(|x_\varepsilon|,|y|)+1$ therefore, $x_\varepsilon,y\in B_R(0)$ and
$ B_R(y)\subset B_{3R}(x_\varepsilon) $
By Mean value property (The mean value is true in any dimension for harmonics functions),
\begin{split} v(y) &=& \frac{1}{|B_R(y)|}\int_{B_R(y)} v(z) dz\\ &=& \frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_R(y)} v(z) dz \\&\le&\frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_{3R}(x_\varepsilon)} v(z) dz \\&= & 3^d v(x_\varepsilon) \end{split} This leads to,
$ v(y)\le 3^d v(x_\varepsilon)\le 3^d(\inf_{\mathbb R^d} v+\varepsilon)~~\forall y\in \mathbb R^d$
That is $ \sup_{\mathbb R^d} v < 3^d \varepsilon $
Since $\inf_{\mathbb R^d} v=0$ ~~ but $\varepsilon>0$ was arbitrarily chosen, letting $\varepsilon \to 0$ we get
$ 0\le \sup_{\mathbb R^d} v \le 0$
i.e $v = 0$ or $u= \inf_{\mathbb R^d} u $