Here is a construction in the $2$-sphere $S^2$, equipped with any reasonable metric. By removing one point it becomes homeomorphic to the plane, so it gives an example in $\mathbb{R}^2$. (You have to be a little careful which point to remove, but it is not that hard to figure out that there exists one that works. Alternatively, equip $\mathbb{R}^2$ with a bounded metric and run the same construction.) The construction is similar to the standard "Lakes of Wada" construction in spirit.
Let $U_1^1$ be a simple path which is $1$-dense in $S^2$, i.e., such that every point on the sphere has distance $\le 1$ to a point on $U_1^1$. Now let $U_2^1$ be a simple path (i.e., a homeomorphic image of $[0,1]$) in $S^2 \setminus U_1^1$ which is $1$-dense in $S^2$. Proceed to get disjoint $1$-dense simple paths $U_1^1,\ldots U_n^1$. Now extend $U_1^1$ to obtain a $1/2$-dense simple path $U_1^2$ in $S^2 \setminus (\bigcup_k U_k^1)$. Inductively construct a sequence of mutually disjoint simple paths $U_1^2,\ldots U_n^2$ which are $1/2$-dense extensions of $U_1^1,\ldots,U_n^1$. Now keep extending those inductively to get mutually disjoint paths $U_1^m,\ldots U_n^m$ which are $1/m$-dense in $S^2$. This construction is possible because at any step the complement of the already constructed paths is connected, since it is the complement in $S^2$ of a finite set of disjoint homeomorphic images of $[0,1]$.
Now let $U_k^\infty = \bigcup_m U_k^m$ for $k=1,\ldots,n$. This is a collection of mutually disjoint open paths (continuous images of $[0,1)$ or $(0,1)$, depending on how exactly the extensions are chosen), each of them dense in the plane. Their union is not necessarily all of $\mathbb{R}^2$, so let $T=S^2 \setminus \bigcup_k U_k^\infty$, and let $U_1 = U_1^\infty \cup T$ and $U_k = U_k^\infty$ for $k\ge 2$. Then $S^2 = \bigcup_k U_k$ is a disjoint partition, and since $U_2,\ldots,U_n$ are continuous images of an interval, they are connected, even path-connected. The set $U_1$ is not necessarily path-connected, so in order to show connectedness assume that $U_1 = A \cup B$ with relatively open disjoint sets $A$ and $B$. Since $U_1^\infty$ is path-connected, it has to be contained in either $A$ or $B$. We may assume $U_1^\infty \subseteq A$. Assume $t \in T \cap B$. Since $U_1^\infty$ is dense and $B$ is relatively open, there has to exist $u \in U_1^\infty \cap B$. However, this contradicts $A \cap B = \emptyset$.
The last argument is probably some standard topology result, that if $U$ is connected, and $V\supseteq U$ is contained in the closure of $U$, then $V$ is connected. The crucial point is to find disjoint connected dense subsets in the first place.
This construction does not guarantee that $U_1$ is path-connected, and I am not sure whether the similar question about a path-connected partition has a positive answer.
The statement is true. Let $X$ be an infinite Hausdorff space. It will suffice to show that we can find a nonempty open set $V_1$ such that $X\setminus\overline V_1$ is infinite. (Then we find $V_2$ in the same way with the infinite Hausdorff space $X\setminus\overline V_1$ playing the role of $X$, and so on.)
At any rate, since $X$ is Hausdorff and has more than one point, we can find a nonempty open set $U$ such that $X\setminus\overline U\ne\emptyset$.
Case 1. If $X\setminus\overline U$ is infinite, let $V_1=U$.
Case 2. If $X\setminus\overline U$ is finite, let $V_1=X\setminus\overline U$. Then $V_1$ is a finite nonempty clopen set, and $X\setminus\overline V_1=X\setminus V_1$ is infinite.
Edit in response to original poster's comment:
If you just wanted to show that for every $n\in\mathbb N$ you can constuct a family of $n$ pairwise disjoint nonempty open sets, you wouldn't need a proof by induction. Just choose $n$ distinct points, choose disjoint neighborhoods for each pair of points, and intersect the (finitely many) chosen neighborhoods of each point. What I described above was a method for constructing an infinite sequence of pairwise disjoint nonempty open sets. I will try to describe the construction a little more formally.
Suppose that $V_1,V_2,\dots,V_n$ have already been defined, so that $V_1,V_2,\dots,V_n$ are pairwise disjoint nonempty open subsets of $X$, and the set $X_n=X\setminus\overline{V_1\cup V_2\cup\dots\cup V_n}=\mathrm{int}(X\setminus(V_1\cup V_2\cup\dots\cup V_n))$ is infinite.
Find a nonempty open set $U\subset X_n$ such that $X_n\setminus\overline U\ne\emptyset$.
If $X_n\setminus\overline U$ is infinite, let $V_{n+1}=U$; if $X_n\setminus\overline U$ is finite, let $V_{n+1}=X_n\setminus\overline U$.
Now $V_1,V_2,\dots,V_n,V_{n+1}$ are pairwise disjoint nonempty open sets, and $X\setminus\overline{V_1\cup V_2\cup\dots\cup V_n\cup V_{n+1}}$ is infinite.
Best Answer
Fix three disjoint dense subsets of $\mathbb{Q}$, call them $D_0, D_1$ and $D_2$.
Let $I_0=(0,1)\backslash \mathbb{Q}$, $I_1=(1,2)\backslash \mathbb{Q}$, $I_2=(2,3)\backslash \mathbb{Q}$ and set $J_k = I_k \cup D_k.$
Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.