Adjoint functors

Given categories $\mathcal{C}$ and $\mathcal{D}$, and functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathbf{U}$ if and only if for every objects $C\in\mathcal{C}$ and $D\in\mathcal{D}$ there is a natural bijection between $\mathcal{C}(C,\mathbf{U}(D))$ and $\mathcal{D}(\mathbf{F}(C),D)$.

Let me use $\leq$ and $\geq$ for the relation among reals, and $\preceq, \succeq$ for the relation among integers.

For $ceil$ to be the left adjoint of the inclusion functor, you would need that for all real numbers $r$ and all integers $z$, $\lceil r\rceil \preceq z$ if and only if $r\leq z$. This holds, so you do have an ajunction (this works because in these categories, the morphism set IntLE$(a,b)$ is empty if $a\not\preceq b$, and contains a unique arrow if $a\preceq b$; and similarly with RealLE; so you get a natural bijection between the sets of arrows if and only if they are either both empty or both are singletons at the same time).

For $floor$ to be a right adjoint to $incl$, you would need that for all real numbers $r$ and all integers $z$, $z\preceq \lfloor r\rfloor$ if and only if $z\leq r$, which again is true; so $floor$ is a right adjoint to the inclusion functor.

For $ceil$ to be the right adjoint to the inclusion functor in the dual categories, you would need $z\succeq \lceil r \rceil$ if and only if $z\geq r$; and for $floor$ to be the left adjoint, you would need $\lfloor r\rfloor \succeq z$ if and only if $r\geq z$. Both hold, so your assertions 1 through 3 are correct.

P.S. Let me second Mariano's suggestion in the comments to keep in mind the case of the underlying set functor and the free group functor for thinking about right and left adjoints. I find myself going back to those two every time I need to remind myself of how things work with adjoints, what adjoints respect or don't respect, and especially when thinking about some of the other equivalent definitions, in particular the one in terms of the unit and co-unit of the adjunction (which are natural transformations between the identity functors and the functors $\mathbf{FU}$ and $\mathbf{UF}$).

Arturo has already posted a nice answer. I'd merely like to emphasize that such universal definitions often enable slick proofs, e.g. see below. For a much more striking example see the theorem in my post here, which presents a slick one-line proof of the LCM * GCD law via their universal definitions.

LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \ $ for $\rm\ \ n > 0$

Proof $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor $

Compare the above trivial proof to more traditional proofs, e.g. the special case $\rm\ m = 1\ $ here.