Let $f $ be a real-valued function defined on a set $S \subseteq \R $.
$f $ is uniformly continuous on $S $ if
$$\forall \epsilon > 0, \exists \delta > 0\text{ s.t. }x, y \in S, \abs{ x - y } < \delta \implies \abs{ f(x) - f(y)} < \epsilon$$
We say $f$ is uniformly continuous if $f$ is uniformly continuous on $\dom{ f } $.
It makes no sense to say a function is uniformly continuous at each point.
If $f $ is continuous on a closed interval $[a, b] $, then $f $ is uniformly continuous on $[a, b] $.
A convergent sequence in $[a, b]$ converges to element in $[a,b] $.
If $f $ is continuous on a closed and bounded set $S $, then $f $ is uniformly continuous on $S $.
If $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence.
A functions $\fnext{ f }$ is an extension of a function $f $ if $\dom{ f } \subseteq \dom{ \fnext{ f }}$and $f(x) = \fnext{ f} (x) $for all $x \in \dom{ f } $.
Let $f $ be a real-valued function on $(a, b)$.
$f $ is uniformly continuous on $(a, b) \iff \exists \fnext{ f } $, $\fnext{ f } $ is continuous on $[a,b] $.
Let $f $ be a continuous function on an interval $I$ ($I$ may be bounded or unbounded). Let $I^{\circ}$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I$.
$f $ is differentiable on $I^{\circ} $ and $f’$ is bounded on $I^{\circ} \implies f $ is uniformly continuous on $I$.