I have written down a proof that real-valued non-constant harmonic functions on $\mathbb{R}^n$ are surjective, but I'm not sure whether my reasoning is valid. So I'd be really appreciated if someone can validate my proof.
Here goes the proof:
Let $u$ be a non-constant harmonic function on $\mathbb{R}^n$. To show that it is surjective, we will show that the image $Y:=u(\mathbb{R}^n)$ is both open and closed.
Let $y\in Y$. Then, there is $x\in\mathbb{R}^n$ such that $u(x)=y$. We can choose a bounded connected neighborhood $V$ of $x$ such that $x$ is neither maximizer or minimizer of $u$ restricted to $V$; otherwise, by Strong Maximum Principle, $u$ would be constant. We have that the image of $V$ contains a neighborhood of $y$ by connectedness of $V$ and continuity of $u$. Since $y$ is arbitrary, $Y$ is open.
Now, let $\{y_k\}\subset u(B_R(0))$, where $B_R(0)\subset\mathbb{R}^n$ is a ball centered at origin with radius $R$, be a sequence converging to $y\in\mathbb{R}$. Then, we have $\{x_k\}\subset B_R(0)$ such that $u(x_k)=y_k$ for each $k$. Since $\{x_k\}$ is in $\overline{B_R(0)}$, which is compact, we must have a convergent subsequence $\{x_{k_i}\}$ converging to some $x\in\overline{B_R(0)}$. So, by continuity, $y_{k_i}=u(x_{k_i})\to u(x)$. Therefore, $y=u(x)\in u(\overline{B_R(0)})\subset Y$. Since $R$ is arbitrary, the image of $u$ is closed.
Finally, since $Y$ is not empty, we conclude that $Y=\mathbb{R}$.
I agree with everything except the last paragraph: it relied only on continuity of $u$ and not that it was harmonic. For instance $u = \exp(x_1)$ works in that last paragraph but it's image is not closed.
One alternative would be this: we know the image is some nonempty interval. Thus if it were not all of $\mathbb{R}$, $u$ would be bounded above or below. But then harmomicity forces $u$ to be constant, which we're assuming it isn't.