Proof of inequality $\lceil x \rceil \le x+1$
The definition of $\lceil x \rceil$ is:

$\lceil x \rceil$ is the minimal integer $n$ such that $n \geq x$.

(The existence of such an integer makes the reals an Archimedean field.)
Let us assume, for the sake of contradiction, that $\lceil x \rceil \geq x + 1$. Then $\lceil x \rceil - 1 \geq x$. Since $\lceil x \rceil - 1$ is also an integer, this contradicts the definition of $\lceil x \rceil$. Thus $\lceil x \rceil &lt; x + 1$.
It is also easy to check that the inequality is tight, in the sense that $1$ cannot be replaced by any smaller $\theta$. Indeed, if $\theta = 1 - \epsilon$ for $\epsilon \in (0,1)$, then $\lceil \epsilon \rceil = 1 = \epsilon + \theta$.