Proof that intervals of the form $[x, x+1)$ or $(x, x+1]$ must contain a integer. Show that any real interval of the form $[x, x+1)$ or $(x, x+1]$ must contain a integer.
Here is my proof (by contradiction)
We start by saying, assume the interval of the form $[x, x+1)$ or $(x, x+1]$ does not contain a integer. Now suppose $x$ is an element of $\mathbb{Z}$, then x is in the interval $[x, x+1)$ and since $x$ is an integer, then $x+1$ must also be an integer.  Thus $(x, x+1]$ must also contain a integer. This is a contradiction, it contradicts our assumption.  Therefore the statement is true.
Can someone help me on this.  Thanks in advance.

You have proved the statement only for $x\in\Bbb Z$.
If $x\notin\Bbb Z$, consider the set $A=\{t\in\Bbb Z: t>x\}$. This set is bounded below and not empty, and it is well-ordered. Let $n=\min A$. Try to prove that $n$ is in the interval.
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