The answer to 1 and 3 should be negative. A space is called $\omega$-bounded if the closure of any countable set is compact. $\omega$-bounded surfaces are well studied (see for instance [Bagpipe theorem][1] and the reference), and an $\omega$-bounded surface $X$ can be written as an increasing union of separable open subsets $\bigcup_{\alpha<\omega_1}X_\alpha$ where the closure of $X_\alpha$ is contained in $X_{\alpha+1}$.

There are many non-homeomorphic $\omega$-bounded surfaces $X$ that can be written as $\bigcup_{\alpha<\omega_1}X_\alpha$ where each $X_\alpha\simeq\mathbb{R}^2$; they are essentially the "long pipes" in the Wikipedia article. Let $L$ be the closed long ray and consider $P=\{(x,y)\in L\times L:x\leq y\}$. The $x$-axis $L\times\{0\}$ and the diagonal $\{(x,y)\in P:x=y\}$ behave very differently. one can show that a continuous map $f:L\rightarrow P$ is either null-homotopic, or homotopic to the inclusion map of $x$-axis, or homotopic to the diagonal map $x\mapsto (x,x)$; in the second case it is either eventually disjoint from the $x$-axis or eventually contained in the $x$-axis, but in the third case it can enter and exit the diagonal unboundedly many times. This can be used to show that the long plane is not homeomorphic to the square of long line. Gluing the diagonal and $x$-axis of $P$ together gives yet another surface, and there are many variation.

Any surface $X$ with the above property is Jordan maximal. It is Jordan because the image of any continuous map from $S^1$ must fall into one of the $X_\alpha$, and is maximal because the image of an $\omega$-bounded space is $\omega$-bounded, and therefore closed (at least when the target space is Hausdorff and first countable, I believe), so the image of an embedding of $\omega$-bounded surface into any surface is clopen, hence equal to that surface.

  [1]: https://en.wikipedia.org/wiki/Bagpipe_theorem