Floor(X) function can be defined as
$\exists$ $M$.$(M< X< M+1 \rightarrow Floor(X)=M) \land (M = X \rightarrow Floor(X)=M)$
Ceiling(X) the function can be defined as
$\exists$ $M$.($M< X < M+1 \rightarrow Ceiling(X)=M+1 \land (M = X \rightarrow Ceiling(X)=M)$
Where $M$ is an integer and $X$ is a real number.
Is there any other better way to decode these functions in First-Order Logic?
The idea is right, but the presentation is a bit sloppy. The key points are:
Definability occurs within a structure (or class of structures), so we need to fix that explicitly.
A definition of a new unary function must have the form $$F(x)=y\quad\iff\quad [stuff],$$ where $[stuff]$ doesn't mention $F$ at all. (And the situation is similar for higher-arity functions, or for relations.)
See this old answer of mine for some background on definability (the question is about set theory specifically, but the answer treats definability in general).
So here's how to correctly phrase what you've done:
Let $\mathcal{A}=(\mathbb{R}; 1, <, +, Z)$, with $1$, $<$, and $+$ having the usual meanings and $Z$ being a unary relation symbol picking out the integers (basically, $\mathcal{A}$ is "the real numbers with the integers labelled").
We want to show that the ceiling function is definable in $\mathcal{A}$. (The situation with the floor function will be the same, and since you have the right idea for both there's no need to treat it separately.) This means that we want to find a formula $\varphi_{ceil}(x,y)$ such that for all $r\in\mathbb{R}$ we have $$\mathcal{A}\models\varphi_{ceil}(r,s)\quad\iff\quad \lceil r\rceil = s.$$ Note that such a $\varphi_{ceil}$ has to be in the language of $\mathcal{A}$ - that is, using only $<$, $+$, and $Z$ - and there may be multiple choices of formula which work here (indeed, there will be).
What you've written in the OP consists of sufficient conditions to establish a value of ceiling which happen to cover all possibilities. We put this in the desired by rephrasing everything in terms of bi-implications: specifically, your idea amounts to the choice $$\varphi_{ceil}\equiv Z(y)\wedge (x=y\vee x<y)\wedge y<x+1.$$ (In English: "$y=\lceil x\rceil$ iff $y$ is an integer and $x\le y<x+1$.") And indeed this formula works.
At this point it's a good exercise to find an analogous choice for $\varphi_{floor}$. In English, we want "$y=\lfloor x\rfloor$ iff $y$ is an integer and $x-1<y\le x$." This is a bit more interesting, since subtraction is not built directly into our language. One option is to change the problem, and work with subtraction explicitly included. However, this is unnecessary, since subtraction itself is definable in terms of $+$. This lets us implement your idea for a definition of $\lfloor \cdot\rfloor$ just in terms of $1,<,+,$ and $Z$ as desired:
In a formula where we want to use subtraction we can introduce a new variable with an auxiliary definition saying that it is the desired difference. For example, the formula $$\varphi_{floor}(x,y)\equiv (y=x\vee y<x)\wedge Z(y)\wedge \forall z[z+1=x\rightarrow z<y]$$ successfully implements your idea. (Note that we could have replaced the "$\forall z$" with an "$\exists z$," since for each $x$ there is a unique $z$ satisfying the hypothesis $z+1=x$.)
However, it turns out that this is overkill. You've used $+$ and $1$, but we don't need that at all: all we need are $<$ and $Z$. Specifically, floor can be defined by
"$\lfloor x\rfloor$ is the biggest integer $\le x$," or more precisely $$\theta_{floor}(x,y)\equiv (y=x\vee y<x)\wedge Z(y)$$ $$\wedge\forall w[(Z(w)\wedge w<x)\rightarrow (w=y\vee w<y)]$$
and ceiling can be defined by
"$\lceil x\rceil$ is the smallest integer $\ge x$," or more precisely $$\theta_{ceil}(x,y)\equiv (x=y\vee x<y)\wedge Z(y)$$ $$\wedge \forall w[(Z(w)\wedge x<w)\rightarrow (y=w\vee y<w)].$$
So really, the optimal result along these lines is:
The floor and ceiling functions are definable in the structure $(\mathbb{R}; <,Z)$ alone.
And for this stronge result, your idea does not work.
