Let $U = \{k\in \mathbb{Z} : x \le k\}$, so $\inf U \ge x$, where $x \in \mathbb{R} $. I want to show that $\inf U < x+1$. Prove this by contradiction.
Suppose $\inf U \ge x+1$. Then, as $\inf U \ge x$, $\inf U \ge \inf U+1$, which is a contradiction.
I don't understand this part. How do we know $\inf U \ge \inf U+1$ from the fact that $\inf U \ge x$?
Thank you in advance.
Edit: I uploaded the image file to clarify question. 
I can't start from $\inf{U} \ge x$ and $\inf{U} \ge x+1$ to get the contradiction you've given. But it seems like Gauss's notation works:
$$ \text{let }\lceil x\rceil = k' \\ $$ Since the infimum of $U$ should not greater than the least element in $U$, $$ \inf{U} \le k' \\ $$ Now since you assumed $\lnot (\inf{U} < x+1),$ $$ x+1 \le \inf{U}\\ x \le \lceil x\rceil < x+1 \le \inf{U} \\ $$
So we have arrived a conclusion that $\inf{U} \le k' < \inf U$, which is a contradiction(You can't find an integer between ... an integer itself).