If $A,B\subseteq X$ where $X$ is a topological space then the pair
$\left\{ A,B\right\} $ is a separation if both sets are not
empty and $\bar{A}\cap B=\emptyset=A\cap\bar{B}$.
If $\left\{ A,B\right\} $ is a separation, and $Y=A\cup B$
then $\left\{ A,B\right\} $ is a separation of $Y$ in
$X$.
Then we have:
$\left\{ A,B\right\} $ is a separation of
$Y$ in $X$ if and only if $\left\{ A,B\right\} $ is a separation of $Y$
in $Y$ .
This affirmes that the definitions are equivalent.
In the following proof $\bar{A}^{Y}$ denotes the closure
of $A$ as a subset of $Y$ and we have $\bar{A}^{Y}=Y\cap\bar{A}$ .
Proof: Let $\left\{ A,B\right\} $ be a separation of $Y$ in $X$.
Then $\bar{A}^{Y}\cap B=Y\cap\bar{A}\cap B\subseteq\bar{A}\cap B=\emptyset$
and likewise $A\cap\bar{B}^{Y}=\emptyset$ showing that $\left\{ A,B\right\} $
be a separation of $Y$ in $Y$. Conversely let $\left\{ A,B\right\} $
be a separation of $Y$ in $Y$. Denoting the complement of $Y$ in
$X$ by $Y^{c}$ we find: $\bar{A}\cap B=\left(Y\cap\bar{A}\cap B\right)\cup\left(Y^{c}\cap\bar{A}\cap B\right)=\left(\bar{A}^{Y}\cap B\right)\cup\emptyset=\emptyset$
and likewise $A\cap\bar{B}=\emptyset$.
Note: if $Y\subset Z\subset X$ then the subspace topology on $Y$ inherited
from $X$ agrees with the subspace topology on $Y$ inherited from
$Z$. So if $\left\{ A,B\right\} $ is a separation of $Y$ in $Z$
then above it has been shown that it is a separation of $Y$ in $X$
as well, and vice versa. So terms like 'in $X$', 'in $Z$' or 'in $Y$'
become redundant here and can be omitted.
You are saying that $A\subseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.
Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).
This comes to the same as proving that no set $U\subset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $U\notin\{\varnothing,f(A)\}$.
Suppose that there is such a set.
Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $V\notin\{\varnothing, A\}$.
Proved is then: $f(A)$ not connected $\implies A$ not connected, or equivalently:
$$A\text{ connected }\implies f(A)\text{ connected}$$
Best Answer
You have the right idea, but it could use elucidation, unless you've got other results to which you can refer.
It seems like you're trying to show that, if $A$ is a clopen subset of $Y$ (as a subspace of $X$) then $A=\emptyset$ or $A=Y,$ and you want to use the fact that this is true when considering $Y$ as a subspace of $S.$ That's a nice idea! Here's how I might go about it.
Suppose that $A$ is a subset of $Y$, and that $A$ is clopen in $Y$ (as a subspace of $X$). We show that $A=\emptyset$ or $A=Y.$
Since $A$ is open in $Y$ as a subspace of $X,$ then there is some $U\in\mathcal T$ such that $A=U\cap Y.$ Since $A$ is closed in $Y$ as a subspace of $X,$ then there is some $V\in\mathcal T$ such that $Y\setminus A=V\cap Y.$
Now, since $A\subseteq Y\subseteq S,$ then $A$ is also a subset of $S,$ and so $$A=U\cap Y=U\cap(S\cap Y)=(U\cap S)\cap Y,$$ so $A$ is open in $Y$ as a subspace of $S.$ We similarly have $Y\setminus A=(V\cap S)\cap Y,$ so $A$ is closed in $Y$ as a subspace of $S.$ Since $Y$ is connected as a subspace of $S,$ then $A=\emptyset$ or $A=Y,$ as desired.