I would like to ask for a clarification to the comments to the question here ;
Proof that every open subset of $\mathbb{R}^n$ is uncountable
As far as I can tell the comment implies for $\textbf{x}\in\mathbb{R}^{n}$ the mapping $f(\textbf{t})\mapsto \textbf{x}+\textbf{t}(1,0,0,…,0)$ for all $|\textbf{t}|<\epsilon$ is a bijective and hence any open set $U\in\mathbb{R}^{n}$ satisfies $|U|=\aleph$. However I could not make this work for me. Instead, letting $e_{1}=(1,0,0,…,0)$,…,$e_{n}=(0,0,0,…,1)$ be the basis vectors for $\mathbb{R}^{n}$ then I believe the mapping $f(\textbf{t})\mapsto \textbf{x}+\sum_{i=1}^{n}\textbf{t}\textbf{e}_{i}$ is in fact required. Where am I going wrong? Below is my "proof" of my claim;
Finally, can the basis vector in the referenced question, or (if I am somehow correct) can the basis vectors in my question, be replaced by basis vectors in a general metric space to come up with a very similar proof – i.e. are all open sets in a metric space uncountable?
************ proof – I switched here to $k$ rather than $n$ *******
For any $\textbf{y}\in\mathbb{R}^{k}$ and $\epsilon>0$ let $B_{\epsilon}(\textbf{y})$ be an open ball centered at $\textbf{y}$. For $i=1,2…,k$ let $\textbf{e}_{i}\in\mathbb{R}^{k}$ be a vector with a $1$ in the $i^{th}$ component and zeroes elsewhere – i.e. $\textbf{e}_{i}$ is the $i^{th}$ basis vector for $\mathbb{R}^{k}$. Now for all $t_{i}$ satisfying $|t_{i}|<k^{-1/2}\epsilon$ we have
\begin{align*}
d_{k}\left[\textbf{y},\textbf{y}+\sum_{i=1}^{k}t_{j}\textbf{e}_{j}\right]&=
d_{k}\left[\sum_{i=1}^{k}y_{i}\textbf{e}_{i},\sum_{i=1}^{k}y_{i}\textbf{e}_{i}+\sum_{i=1}^{k}t_{i}\textbf{e}_{i}\right]\\
&=\sqrt{\sum_{i=1}^{k}\left(y_{i}(\textbf{e}_{i})_{i}-y_{i}(\textbf{e}_{i})_{i}-t_{i}(\textbf{e}_{i})_{i}\right)^{2}}\\
&=\sqrt{\sum_{i=1}^{k}\left(-t_{i}\right)^{2}}\\
&=\sqrt{\sum_{i=1}^{k}t_{i}^{2}}\\
&<\sqrt{k\cdot k^{-1}\epsilon^{2}}\\
&=\epsilon.
\end{align*}
Accordingly for $\textbf{t}:=(t_{1},…,t_{k})$ satisfying $|t_{i}|<k^{-1/2}\epsilon$ for all $i=1,…,k$ (which implies $||\textbf{t}||_{k}=\sqrt{\sum_{i=1}^{k}t_{i}^{2}}<\sqrt{k\cdot k^{-1}\epsilon^{2}}=\epsilon$), if we define the vector $\textbf{y}_{\textbf{t}}=\textbf{y}+\sum_{j=1}^{k}(\textbf{t})_{j}\textbf{e}_{j}$ then the above equation means $d_{k}[\textbf{y},\textbf{y}_{\textbf{t}}]<\epsilon$ which implies $\textbf{y}_{\textbf{t}}\in B_{\epsilon}(\textbf{y})$. Thus defining the mapping $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$, $f_{\textbf{y}}(\textbf{t})\mapsto \textbf{y}_{\textbf{t}}$, we have $f_{\textbf{y}}((-\epsilon,\epsilon))\subseteq B_{\epsilon}(\textbf{y})$. Conversely choose a $\textbf{x}\in B_{\epsilon}(\textbf{y})$ which implies $d_{k}[\textbf{x},\textbf{y}]:=d<\epsilon$, and write $\textbf{x}$ and $\textbf{y}$ in terms of the basis vectors $\textbf{x}=\sum_{i=1}^{k}\textbf{x}_{i}\textbf{e}_{i}$ and $\textbf{y}=\sum_{i=1}^{k}\textbf{y}_{i}\textbf{e}_{i}$ so that $d_{k}^{2}[\textbf{x},\textbf{y}]$ can be written as
\begin{align*}
d_{k}^{2}[\textbf{x},\textbf{y}]&=\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)^{2}(\textbf{e}_{i})_{i}^{2}\\
&=\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)^{2}\\
&=d^{2}\\
&<\epsilon^{2}.
\end{align*}
Defining $t_{i}(\textbf{x},\textbf{y}):=(\textbf{x}_{i}-\textbf{y}_{i})$, and in turn $t(\textbf{x},\textbf{y})=\left(t_{i}(\textbf{x},\textbf{y})\right):=((\textbf{x}_{1}-\textbf{y}_{1}),…,(\textbf{x}_{k}-\textbf{y}_{k}))$, using the above equation implies $||t(\textbf{x},\textbf{y})||_{k}=d_{k}[\textbf{x},\textbf{y}]=d<\epsilon$. Thus we have
\begin{align*}
\textbf{x}&=\textbf{y}+\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)\textbf{e}_{i}\\
&=\textbf{y}+\sum_{i=1}^{k}t_{i}(\textbf{x},\textbf{y})\textbf{e}_{i}\\
&=\textbf{y}+\textbf{t}(\textbf{x},\textbf{y})\textbf{e}_{i}\\
&=\textbf{y}_{\textbf{t}(\textbf{x},\textbf{y})},
\end{align*}
for some $||t(\textbf{x},\textbf{y})||_{k}<\epsilon$, which leads to the conclusion $\textbf{x}\in f_{\textbf{y}}((\epsilon,\epsilon))$. Accordingly $B_{\epsilon}(\textbf{y})\subseteq f_{\textbf{y}}((-\epsilon,\epsilon))$ and so the conclusion $f_{\textbf{y}}((-\epsilon,\epsilon))= B_{\epsilon}(\textbf{y})$ follows. Thus $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$ is an onto function. Furthermore for all $\textbf{s},\textbf{t}\in\mathbb{R}^{k}$ since $(\textbf{y})_{i}+(\textbf{s})_{i}=(\textbf{y})_{i}+(\textbf{t})_{i}$ if and only if $(\textbf{s})_{i}=(\textbf{t})_{i}$ for all $i=1,…,k$ implies $\textbf{y}+\sum_{j=1}^{k}(\textbf{s})_{j}\textbf{e}_{j}=\textbf{y}+\sum_{j=1}^{k}(\textbf{t})_{j}\textbf{e}_{j}$ if and only if $\textbf{s}=\textbf{t}$ then $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$ is also an injective function, and hence is a bijection. Now $|(-\epsilon,\epsilon)|=\aleph$ is well known [and I do not prove it here] and so $|-\epsilon,\epsilon|=|B_{\epsilon}(\textbf{y})|$ (i.e. $(-\epsilon,\epsilon)$ is equipotent to $B_{\epsilon}(\textbf{y})$). By Definition of equality of set cardinalities we conclude $|(-\epsilon,\epsilon)|=| B_{\epsilon}(\textbf{y})|=\aleph$.
Best Answer
Taking onboard all the comments above, it seems every open ball in $\mathbb{R}^{k}$ is uncountable [proof below as per the original proposition in the linked question], and since all non-empty open sets in $\mathbb{R}^{k}$ are a union of open balls, and unions of uncountable sets produce another uncountable set, then we conclude all non-empty open sets in $\mathbb{R}^{k}$ are uncountable.
Finally it has been shown by examples that in general open sets in metric spaces need not be uncountable.
Proof that all open balls in $\mathbb{R}^{k}$ are uncountable;
For any $\textbf{y}\in\mathbb{R}^{k}$ and $\epsilon>0$ let $B_{\epsilon}(\textbf{y})$ be an open ball centered at $\textbf{y}$. Let $\textbf{e}_{1}\in\mathbb{R}^{k}$ be the vector with a $1$ in the $1^{st}$ component and zeroes elsewhere - i.e. $\textbf{e}_{1}$ is the $1^{st}$ basis vector for $\mathbb{R}^{k}$. Now for all $t$ satisfying $|t|<\epsilon$ we have
\begin{align*} d_{k}\left[\textbf{y},\textbf{y}+t\textbf{e}_{1}\right]&= d_{k}\left[\sum_{i=1}^{k}y_{i}\textbf{e}_{i},\sum_{i=1}^{k}y_{i}\textbf{e}_{i}+t\textbf{e}_{1}\right]\\ &=\sqrt{\sum_{i=1}^{k}\left(y_{i}(\textbf{e}_{i})_{i}-y_{i}(\textbf{e}_{i})_{i}-t(\textbf{e}_{1})_{i}\right)^{2}}\\ &=\sqrt{\left(-t(\textbf{e}_{1})_{1}\right)^{2}}\\ &=\sqrt{t^{2}}\\ &=|t|\\ &<\epsilon. \end{align*}
Accordingly if we define the vector $\textbf{y}_{t}=\textbf{y}+t\textbf{e}_{1}$ then the above equation means $d_{k}[\textbf{y},\textbf{y}_{t}]<\epsilon$ which implies $\textbf{y}_{t}\in B_{\epsilon}(\textbf{y})$. Thus defining the mapping $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$, $f_{\textbf{y}}(t)\mapsto \textbf{y}_{t}$, we have $f_{\textbf{y}}((-\epsilon,\epsilon)):=S\subseteq B_{\epsilon}(\textbf{y})$.
Furthermore for all $s,t\in\mathbb{R}$ since $\textbf{y}+s\textbf{e}_{1}=\textbf{y}+t \textbf{e}_{1}$ if and only if $s=t$ then $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow S$ is an injective and obviously onto function, and hence is a bijection. Now $|(-\epsilon,\epsilon)|=\aleph$ is well known [and I do not prove it here] and so by the definition of comparing cardinalities of sets we have $\aleph=|-\epsilon,\epsilon|\leq |B_{\epsilon}(\textbf{y})|$. But since $|B_{\epsilon}(\textbf{y})|<|\mathbb{R}^{k}|=\aleph$ [proof omitted but using bijective functions again] we conclude $\aleph=|B_{\epsilon}(\textbf{y})|$.
Proof that all open sets in $\mathbb{R}^{k}$ are uncountable;
From the above proof and the definition of comparing cardinalities of sets using bijective functions, the definition of the open sets $\mathcal{U}(\mathbb{R}^{k})$ implies for any $U\in\mathcal{U}(\mathbb{R}^{k})$ that $\aleph\leq |U|$, for choosing any $u\in U$ there exists a $\epsilon_{u}$ such that $B_{\epsilon_{u}}(u)\subseteq U$ where $|B_{\epsilon_{u}}(u)|=\aleph$ which means the bijection $\phi:B_{\epsilon_{u}}(u)\longrightarrow B_{\epsilon_{u}}(u)\subseteq U$, $\phi(u)\mapsto u$, injects a set with cardinality of the continuum into $U$. Furthermore since $|\mathbb{R}^{k}|=\aleph$, the fact that $U\subset\mathbb{R}^{k}$ and the existence of the bijection $\phi:U\longrightarrow U\subseteq \mathbb{R}^{k}$, $\phi(u)\mapsto u$ combined the definition of comparing set cardinalities implies $|U|\leq\aleph$. Thus we have $|U|=\aleph$