A uniformly continuous function maps bounded set to bounded sets

Assume that $f$ is unbounded, and $\sup_{x\in A} f(x)=\infty$. (The case $\inf_{x\in A} f(x)=-\infty$ can be treated in the same way.)

Then, there is a sequence $\{x_n\}\subset A$, such that $f(x_n)\to\infty$. We can pick a subsequence $\{y_n\}$ of $\{x_n\}$, such that $f(y_{n+1})-f(y_n)>1$, for all $n\in\mathbb N$. Since $f$ is uniformly continuous, there exists a $\delta>0$, such that, for all $x,y\in A$, $$ |x-y|<\delta\quad\Longrightarrow\quad |\,f(x)-f(y)|<1.\tag{1} $$ But as $A\subset\mathbb R$ is bounded, then $\{y_n\}$ has a convergent subsequence $z_n\to z\in \overline{A}$. In fact, we may pick the subsequence $\{z_n\}$, so that $|z_m-z_n|<\delta$, for all $m,n\in\mathbb N$, which implies that, for all $m,n\in\mathbb N$, with $m\ne n$, we have $$ |z_m-z_n|<\delta, \quad\text{while}\quad |f(z_m)-f(z_n)|>1, $$ which contradicts $(1)$.

Note. Since $f$ is uniformly continuous, then $f$ extends continuously to $\overline{A}$. This is done in the following way. If $\{x_n\}\subset A$ is Cauchy, then so is $\{f(x_n)\}$, due to the uniform continuity of $f$. Hence, $f(x)$ can be extended continuously (and uniquely) to $\overline{A}$.

But $\overline{A}$ is compact, since $A$ is bounded. The extension of $f$ shall be bounded, as it is continuous on a compact set, and thus $f$ is bounded on $A$.