I have attempted this proof and would appreciate any feedback. Thanks.

Let $\displaystyle I$ be some interval. If $\displaystyle f:I\rightarrow \mathbb{R}$ is increasing and $\displaystyle a\in I$, prove that $\displaystyle f(a)$ is an upper bound for $\displaystyle \{f(x)|x<a\}$ and $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.

Proof: Let $\displaystyle f$ be an increasing function. Then, by definition, for all $\displaystyle x,y\in I, x<y\Rightarrow f(x)<f(y)$. Then, let $\displaystyle x<a$ for $\displaystyle a\in I$ and all $\displaystyle x\in I$. Thus, since $\displaystyle x<a\Rightarrow f(x)<f(a)$ for all $\displaystyle a,x\in I$. Hence, $\displaystyle a$ is an upper bound for $\displaystyle \{f(x)|x<a\}$.

Then, since $\displaystyle x<a$ for all $\displaystyle x,a\in I$, then $\displaystyle a$ must be the least upper bound for $\displaystyle \{f(x)|x<a\}$. Hence, $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.