Can we partition an open set by finitely many open subsets? If we have open set, e.g. $(0, 1)$ or an open disc, is it possible to partition the it by finitely many open subsets?
Obviously we ignore the trivial case which is the subset itself.
 A: Let $X$ be a topological space. We say $X$ is disconnected if there are disjoint open non-empty subsets $A, B \subset X$ such that $X = A\cup B$. If $U$ is not disconnected, we say that it is connected.
Suppose $X$ is an open set which can be partitioned into finitely many proper open subsets. Let $X = \bigcup_{i=1}^nU_i$ where the $U_i$ are disjoint open non-empty sets (and therefore proper subsets of $X$). Setting $A = U_1$ and $B = \bigcup_{i=2}^nU_i$, we see that $X = A\cup B$ where $A$ and $B$ are disjoint open non-empty subsets of $X$. Therefore, any such $X$ must be disconnected. It is worth noting that we don't need to assume the partition is finite. That is, we have the following stronger statement: 

If $X$ can be partitioned by proper open subsets, then $X$ is disconnected.

As $(0, 1)$ and the open disc $\{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < r\}$ are connected, the above result shows that they cannot be partitioned into finitely many proper open sets.
Any disconnected space $X$ can be partitioned into its connected components $\{C_i \mid i \in I\}$ which are closed. If $X$ has finitely many connected components (i.e. $I$ is finite), then the connected components are also open. Then the union $X = \bigcup_{i\in I} C_i$ provides a partition of $X$ into finitely many proper open subsets. Note, the assumption on the number of connected components is necessary as the connected components of a disconnected space can need not be open. For example, the connected components of $\mathbb{Q}$ are singletons which are closed but not open.
