Is $\sigma(S^c) = \mathcal{P}(X)$, where $S^c$ is a complement of an arbitrary sigma algebra on X? Is the smallest sigma-algebra containing the complement of an arbitrary sigma algebra $S$ on $X$ always equal to the power set $\mathcal{P}(X)$? In short, does $\sigma(S^c) = \mathcal{P}(X)$?
By complement of a sigma algebra I mean $\mathcal{P}(X) \setminus S$.
EDIT: except the trivial case when $S = \mathcal{P}(X)$
(Definitely, for a non-trivial sigma-algebra, meaning $|S| > 2, S \ne \mathcal{P}(X)$, we can't say the opposite, that is "never equal to the power set". Counterexample, when $\sigma(S^c)$ is actually the power set -> $S = \{\{\}, \{1,2,3\}, \{1\}, \{2, 3\}\},X = \{1,2,3\}$.)
Or we can't say anything at all about such $\sigma(S^c)$?
 A: What about
$$S = \{A \subset \mathbb{R} | A \textit{ countable or } A^c \textit{ countable}\}$$
Then $$S^c = \{A \subset \mathbb{R} | A \textit{ uncountable and } A^c \textit{ uncountable}\}$$
How would $$ \sigma(S^c)$$ include any infite, countable sets?
A: This was supposed to be a proof, but in the proof of the Lemma, there is a problem described under "Recent realization".
However, if the Lemma holds, I think the rest of the theorem proof holds (but I don't have time now to double check it).
Theorem in question
Let $S$ be a $\sigma$-algebra on an arbitrary set $X$, s.t. $S \ne \mathcal{P}(X)$. Then $\sigma(S^c) = \mathcal{P}(X)$.

First, inspired by why the situation in the first answer doesn't work, we'll show, that:
Lemma
(i) A  $\sigma$-algebra on an arbitrary set $X$ $$S = \{A \subseteq X | A \textit{ countable or } A^c \textit{ countable}\}$$
or (ii) any $\sigma$-algebra generated by superset of $S$, meets the theorem in the question.
Proof
If $X$ is countable, then the theorem's premise $\sigma(S) \ne \mathcal{P}(X)$ doesn't hold, resp. the theorem holds. Now for $X$ uncountable:
Take $S$. $S^c$ is non-empty. For any $x \in X$, take a set $A \in S^c$. Then $x \in A \text{ or } x \in A^c$. WLOG, let $x \in A$. $A \setminus \{x\}$ is also uncountable therefore in $S^c$. $(A\setminus\{x\} \cup A^c)^c = \{x\} \in \sigma(S^c)$. Because $\sigma(S^c)$ is a $\sigma$-algebra, there are all countable unions of any such one-element sets, therefore there are all countable sets. So $\sigma(S^c) = \mathcal{P}(X)$. So Lemma holds for $S$.
To avoid getting this result, we would need to exclude from $\sigma(S^c)$ at least one set.
EDIT: the following part is too problematic, I think. I should formalize that. What does the second sentence mean? And what if there is no such $i$, because you removed all countable sets.
First, let the excluded set $C$ be countable. Then we need to exclude from $\sigma(S^c)$, actually from $S^c$, at least, always one of the member of those uncountable set pairs $A$, $B \in S^C$ where $A \cap B = \{i\}$, for some $i \in X$. (Otherwise we could create any countable set, as seen earlier.) Therefore adding them to $S$. But then $S$ becomes $\mathcal{P}(X)$ (which is a contradiction), because originally, for every $A \in S^c$, we need to add $A$ or $A^c$ (depending in which $i$ is) to $\sigma(S)$. WLOG $i \in A$, and for every such $A$ there existed such $B = A^c \cup \{i\}$. ($B$ and $B^c$ are uncountable, therefore in $S^c$.) So even $\sigma$-algebra of superset of $S$ cannot avoid being equal to the power set, if we were to remove a countable set from $\sigma(S^c)$.
Recent realization:
Unfortunately, we still can try to remove an uncountable set from $\sigma(S^c)$. But I don't have a proof that it would give the same result, yet, i.e. that $\sigma$-algebra of the superset of $S$ was the power set. Or maybe it is not true. Then this could be the place to disprove the theorem by finding an example -> find how to exclude at least one uncountable set from $\sigma(S^c)$, while keeping $\sigma(S)$ smaller than the power set.
Almost '$\square$'
A simple observation is also that for any subset of $S$ the Lemma holds too. (Because if $R \subset S$, $\sigma(R^c) \supseteq \sigma(S^c) = \mathcal{P}(X)$.)
Proof of the theorem in question
If $S$ contains all countable sets, then the theorem holds according to the Lemma.
If $S$ contains none of the countable sets, then $S^c$ contains all of them, therefore also $\sigma(S^c)$, and plugging it in the Lemma, we get contradiction $S = \mathcal{P}(X)$
Now it suffices to show the theorem holds if $S$ contains at least one countable set and $S^c$ contains at least one countable set, say set $B$. Note it also means $B^c \in S^c$, because if it weren't $B$ was also in $S$, which is a contradiction.
We will show that whenever $S$ contains a one-element set $I=\{i\}$, $\sigma({S^c})$ also contains it, therefore when applying the Lemma for $\sigma$-algebra $\sigma({S^c})$, we get contradiction that $S = \mathcal{P}(X)$.
Focus on the (arbitrary) one-element set $I \in S$. Necessarily, there is set $B^{any} \in \{B, B^c\}, i \in B^{any}$. Set $B^{any} \setminus I$ is either in $S^c$ and we're done, as $I = ((B^{any})^c \cup (B^{any} \setminus I))^c \in \sigma(S^c)$, or it has to be in $S$. But then we get a contradiction -> $B^{any} = B^{any} \setminus I \cup I$  has to be in $S$ and it was in $S^c$
We have shown every one-element set is in $S^c$, therefore every countable set is in $\sigma({S^c})$. After applying the Lemma for $\sigma({S^c})$,  we get contradiction that $S = \mathcal{P}(X)$. $\square$
A: After all, I think I have a simple counterexample.
Take $$S = \sigma(\{ A \subset \left[ 0,1\right] | A \textit{ countable }\}  \cup \{A \subset \mathbb{R} \setminus \left[ 0,1 \right] \} )$$
Then both $S$ and $\sigma(S^c)$ are not $\mathcal{P}(\mathbb{R})$, right?
