A subspace $Y$ of a topological space $X$ is disconnected exactly when it can be written as a union $Y=A\cup B$ with $A$ and $B$ disjoint nonempty open subsets of $Y$ (in the subspace topology). In terms of the topology of $X$, we have $A=U\cap Y$ and $B=V\cap Y$ for open sets
$U$ and $V$ in $X$. This translates to:
$(1)$ $Y$ is disconnected iff $Y\subset U\cup V$ for some open sets $U$ and $V$ in $X$, both meeting $Y$, and with $U\cap Y$ and $V\cap Y$ disjoint.
As indicated in Willard, Exercise 26D, the following is not true in general:
$(2)$ $Y$ is disconnected iff $Y\subset U\cup V$ for some disjoint open sets $U$ and $V$ in $X$, both meeting $Y$.
We cannot take $U$ and $V$ disjoint in condition $(1)$ in general. See the answer to this question for a $T_0$ example. A simple $T_1$ example is the line with two origins $X=\mathbb{R}\cup\{0^*\}$. The subspace $Y=\{0,0^*\}$ is disconnected, with the only possible disconnection being $\{0\}\cup\{0^*\}$, but any two respective nbhds of $0$ and $0^*$ in $X$ are not disjoint.
However, if $X$ is a metric space, we can require $U$ and $V$ to be disjoint, as shown here.
Question:
Is there an example of a Hausdorff space $X$ with a disconnected subspace $Y$, where the disconnection cannot be witnessed by the traces on $Y$ of disjoint open sets in the ambient space?
As mentioned above, any counterexample cannot be a metrizable space. Also $Y$ cannot be finite, as any two points can be separated by disjoint open sets in a $T_2$ space. I just can't come up with any example.
Best Answer
With respect to the usual topology on $\Bbb R^2,$ for $n\in \Bbb N$ let $C_n$ be the closed semi-circular arc in $\Bbb R\times [0,\infty)$ joining $(1/n,0)$ to $(1/(n+1),0).$
Let $X=(\,[0,\infty)\times \{0\}\,)\cup (\cup_{n\in\Bbb N}C_n\,).$
Let $T$ be the usual topology on $X$ as a subspace of $\Bbb R^2$. Let the topology $T_X$ be the weakest topology $T'$ on $X$ such that $T'\supset T$ and such that $\cup_{n\in\Bbb N}C_n$ is $T'$-closed. Let $C=\cup_{n\in\Bbb N}C_n$ and let $D=\{(0,0)\}$ and let $Y=C\cup D.$
Details are for the reader. Hints: (1). $C$ and $D$ are connected closed subspaces of $X$. (2). If $U,V\in T_X$ with $C\subset U$ and $D\subset V$ then $(U\cap V)\cap (X\setminus Y)\ne \emptyset.$