# Show that $x \leq a$.

Let $a,x \in [-\infty , \infty)$. Assume that for each number $c$ such that $c>a$ we have the inequality $x\leq c$. Show that then $x\leq a$.

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Suppose $x>a$. Take a number $b$ in the interval $(a,x)$. So $b>a$ and by hypothesis $x\leq b$. Absurd.

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A proof by contradiction should start by assuming that $x\not\le a$, i.e., that $x>a$. Let $c=\frac12(x+a)$, and derive a contradiction.
Take $A=\{c \in \mathbb R | c> a\}$. Clearly, a is the infimum of A. On the other hand x is a lower limit for A. This implys $x \le a$.