There are two things you can do with Topology: (1) try to imagine what's going on intuitively, and (2) write it down rigorously. Both are useful, necessary steps.
For instance, if I understand your first example, you have the unit square $[0,1]^2 \subset \mathbb{R}^2$ with the subspace topology. Let me write the coordinates of its points $(\theta, z) \in [0,1]^2$, for psychological reasons that will become apparent in a minute.
On the intuitive side, what you are doing is bending your square $[0,1]^2$ in such a way that sides $\theta = 0$ and $\theta = 1$ become closer and closer... You put some glue on them, stick them... Ok: and what do you "see" now? -A cylinder, right?
(Except for the fact that in real life you cannot stick just two segments -or it's really difficult-, you can try to do this experiment with a piece of paper and some glue.)
On the rigorous side, you may do the following: you recover that parametrization of the cylinder you have probably seen in some geometry or integral calculus course
$$
\varphi: [0,1]^2 \longrightarrow \mathbb{R}^3 \ , \qquad \varphi (\theta , z) = (\cos (2\pi\theta ), \sin (2\pi\theta ), z) \ .
$$
This $\varphi$ is obviously continuous ("obviously" means: "someone taught you this in some calculus course a year ago") and surjective, if we restrict the codomain to the cylinder
$$
C = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 = 1 \ , \ z \in [0,1] \right\} \ .
$$
That is, we are looking to $\varphi$ as a map $\varphi : [0,1]^2 \longrightarrow C$. Notice that also this way it is continuous with the subspace topologies on both sides.
Now $\varphi $ is almost injective too, except for those points $(0,z)$ and $(1,z)$ that have the same image
$$
\varphi (0,z) = (1,0,z) = \varphi (1,z) \ , \qquad \text{for all}\qquad z \in [0,1] \ .
$$
In order to fix this lack of injectivity, you say: "Ok, let's identify ("identify" is the picky, correct, way to say "glue" in Topology) each $(0,z)$ with its corresponding $(1,z)$ for all $z$." So, you are defining an equivalence relation $\sim$ among the points of the square $[0,1]^2$, the one generated by:
$$
(0,z) \sim (1,z) \qquad \text{for all} \qquad z \in [0,1] \ .
$$
You call $X = [0,1]^2/\sim$ the space obtained from the square $[0,1]^2$ after gluing (sorry, "identifying") each $(0,z)$ with its corresponding $(1,z)$.
Let's write $\widetilde{(\theta ,z)} \in X$ the equivalence classes produced by this equivalence relation. Notice that, if $\theta \neq 0, 1$, you just have $\widetilde{(\theta , z)} = \left\{ (\theta , z) \right\}$, but $\widetilde{(0,z)} = \widetilde{(1,z)} = \left\{ (0,z), (1,z) \right\}$. That is, points of $X$ are "the same" as the points of $[0,1]^2$, except for those $(0,z)$ and $(1,z)$ that are now glued togehter ("identified", yes).
In doing this, we also get a natural map ("projection", "identification")
$$
\pi : [0,1]^2 \longrightarrow X \ , \qquad \pi (\theta , z) = \widetilde{(\theta , z)} \ ,
$$
which is continuous by definition of the quotient topology on $X$.
Back to our $\varphi$: since it identifies the same points as $\sim$ does, $\varphi$ induces a well-defined map
$$
\widetilde{\varphi} : X \longrightarrow C \ , \qquad \widetilde{\varphi}\widetilde{(\theta , z)} = \varphi (\theta ,z) \ .
$$
Now, since we have glued the points with the same image by $\varphi$, this new $\widetilde{\varphi}$ is bijective, but it's also continuous, because of the universal property of the quotient topology and the fact that $\widetilde{\varphi} \circ \pi = \varphi$. Right?
Last step. A continuous bijective map doesn't need to be a homeomorphism, as you probably know: the inverse $\widetilde{\varphi}^{-1}$ is not necessarily continuous.
What could we do? Well, you can try to write down an explicit formula for $\widetilde{\varphi}^{-1}$ and check its continuity directly, but this is hard and painful, so nobody does it.
Instead, everybody resorts to the following marvelous, great, second to none trick (the best one you can buy in elementary Topology; btw, "trick" means "proposition": you can prove it, of course):
"$X$ is compact, because it is a quotient of a compact space (namely, $[0,1]^2$). $C$ is Hausdorff, because it is a subspace of a Hausdorff space (namely, $\mathbb{R}^3$).
The GTET (Greatest Trick in Elementary Topology)
Now, I have a continuous, bijective map $\widetilde{\varphi} : X \longrightarrow C$ between a compact space and a Hausdorff one. Hence $\widetilde{\varphi}$ is a homeomorphism."
Isn't that fantastic? :-) (Don't forget it the next time you need to prove that some map is a homeomorphism: it will save your life.)
So, you can also prove rigorously your first intuition: your space is a cylinder, $X \cong C$.