Hint.
First prove that if $K$ is any component of $(\mathbb{R}^n-S)$, then $S\cup K$ is connected.
Some details:
It is easily seen that $K$ is closed. It cannot also be open as this would partition $\mathbb{R}^n$. If $p$ is any boundary point of $K$ then $p\in\overline{S}$, for otherwise we could add a small ball around $p$ and obtain a strictly bigger than $K$ connected set missing $S$, a contradiction. It follows that $S\cup\{p\}$ and $K$ are two connected sets with a non-empty intersection, hence their union is connected.
Next represent $(\mathbb{R}^n-T)$ as the union of a family of connected sets that do have a non-empty intersection:
Indeed $(\mathbb{R}^n-T)=\bigcup\{S\cup K: K$ is a component of $(\mathbb{R}^n-S)$ and $K\not=T \}$ .
Edit.
We do not need to assume that $S$ is open.
If $S$ need not be open, then $K$ (in the notation introduced above) need not be closed in $\mathbb{R}^n$. But, just as before, $K$ cannot be both closed and open, hence it has a non-empty boundary $\mathrm{Bd\,}K$. Pick $p\in\mathrm{Bd\,}K$. If $p\in S$ then $S$ and $K\cup\{p\}$ are two connected sets with a non-empty intersection, hence their union $S\cup K$ is connected. If $p\not\in S$, then $p\in K$ (since $K\cup\{p\}$ is connected and $K$ is a component of $(\mathbb{R}^n-S)$, i.e. a maximal connected subset). As before we must have that $p\in\overline S$, and that $S\cup\{p\}$ and $K$ are two connected sets with a non-empty intersection, hence their union $S\cup K$ is connected.
The above may be generalized as follows.
Suppose that $X$ is a connected topological space and $S$ is a connected subset. Suppose also that $\overline S\cap \overline K\not=\emptyset$ for every component $K$ of $X-S$. Then, if $T$ is any component of $X-S$, we have that
$X-T$ is connected.
The condition above that $\overline S\cap \overline K\not=\emptyset$ certainly holds if $X$ is locally connected. Indeed, if $\overline S\cap \overline K=\emptyset$ then $\overline K$ is a connected set missing $S$, hence $K=\overline K$, i.e. $K$ is closed. But $K$ is also open (as a component) in $X-\overline S$ since $X-\overline S$ is locally connected (being open in the locally connected $X$). Then $K$ is both closed and open in $X$, a contradiction.
I do not know if the condition that $X$ is locally connected is necessary. Perhaps $\overline S\cap \overline K\not=\emptyset$ always holds, regardless of whether $X$ is locally connected or not? I posted this as a separate question .
Edit. My separate question was answered (with a link to the answer of another older question). There is indeed a space $X$ (that is not locally connected at just two points), a connected subset $S$ (which in that example is a singleton $\{(0,1)\}$, but could be made open by taking a small neighborhood), and a component $K$ of $X\setminus S$ (namely $K=\{(0,0)\}$ again a singleton) such that
$\overline S\cap \overline K=\emptyset$.
Question (which I will not post for now as it is getting too late, please feel free to post separately if you wish). Assume that $X$ is a connected topological space, $S$ is a connected subset, and $T$ is a component of $X\setminus S$. Is $X\setminus T$ connected? My particular proof here does not work in general, but perhaps the answer is nevertheless yes, with another proof?
