I have to prove that for some uncountably infinite set $ S$ and $x \in S$ the cardinality of $S$ is equal to the cardinality of $S \setminus \{x\}$.
I have proven that this is the case for countably infinite sets but I cannot figure out how to prove if for uncountable sets.
It doesn't matter the set is countably infinite or uncountable, the result is true for every infinite set. And not only single element, one can remove a finite set of elements from that infinite set without changing it's cardinality.
$S$ be an infinite set and $A \subset S$ be any finite set.
Claim : $Card(S)=Card(S\setminus A)$
Proof: Since $A$ is finite, assume $$A\sim\mathbb{N_n}$$ for some $n\in \Bbb{N}$.
Let, $A=\{x_1,x_2,,...,x_n\}$
Then by Axiom of choice, there exists a choice function
$$f: {\scr{P}}(S)\to S $$
such that $f(E) \in E \space$, $\emptyset \neq E\subset X$
Then define,
\begin{align}x_{n+k} &=f(S\setminus \{x_1,x_2,...,x_{k}\})&k=1,2,... \end{align}
Now, define a map $$h:S\to S\setminus A$$ by
\begin{equation} h(x) = \begin{cases} x_{n+k} & \text{; } x=x_k\\ x & \text{ otherwise } \end{cases} \end{equation}
Then, $h$ define a bijection.
Hence, $Card(S) =Card(S\setminus A) $
In your question the set $A$ is a singleton set. The proof is same. Try by yourself. Thanks.