Find $f:C\to\mathbb{R}^2$ continuous and bijective but not open, $C\subset\mathbb{R} ^2$ is closed and connected Are there a closed connected subspace  $C$ of $\mathbb{R}^2$ and a continuous, bijective function  $f:C\to\mathbb{R}^2$ that is not open?
If we remove the condition for $C$ to be connected, we have the function 
$f:\big([0,+\infty)\times\mathbb{R}\big)\cup\{(-1,0)\}\to\mathbb{R}^2$ defined by
\begin{equation}
f(x,y)=
\begin{cases}
&\bigg(e^y\cos\big(\frac{2\pi x}{x+1}\big),e^y\sin\big(\frac{2\pi x}{x+1}\big)\bigg)&,&(x,y)\neq (-1,0)\\
&(0,0)&,&(x,y)=(-1,0)
\end{cases}  
\end{equation}
which satisfy the rest of the conditions. See justification here.
But what is the answer if $C$ is closed and connected? 
 A: 
Are there a closed connected subspace  $C$ of $\mathbb{R}^2$ and a continuous, bijective function  $f:C\to\mathbb{R}^2$ that is not open?

No. Suppose $C\subseteq \mathbb R$ is closed and connected. Then either 
(0) $C=\varnothing$,
(1) $C\simeq \{0\}$,  
(2) $C\simeq [0,1]$, 
(3) $C\simeq [0,\infty)$, or 
(4) $C\simeq \mathbb R$.

(0) and (1): Cannot work because in this case $C$ and $\mathbb R^2$ have different cardinalities. 
(2): Cannot work because $[0,1]$ is compact but $\mathbb R ^2$ is not.
(3): Suppose $f:[0,\infty)\to \mathbb R ^2$ is continuous and injective. Then for each $n\in\omega$, $f\restriction [0,n]$ is a homeomorphism, as a continuous function from a compact space to a Hausdorff space is closed. Thus $f[0,n]$ is closed and nowhere dense in $\mathbb R ^2$ (if it had interior then the image would have non-cut points, whereas $[0,n]$ has no cut points).Thus we find $f[0,\infty)$ is a countable union of closed nowhere dense subsets of $\mathbb R ^2$, which cannot be all of $\mathbb R ^2$ by the Baire Category Theorem for complete metric spaces. So $f$ is not surjective.
(4): Replace $[0,n]$ with $[-n,n]$ in the argument above.
EDIT: I now see that there was a typo in the title of the thread. I guess $C$ should be a subset of $\mathbb R ^2$ not $\mathbb R$. Nevertheless, I have ruled out some possibilities for $C$.
