Uniform continuity on $[c,d] \subseteq (a,b)$ implies UC on $(a,b)$ and UC on $[a,b] \ \forall a,b \in \mathbb{R}$ implies UC on $\mathbb{R}$ Is it true that if $f$ is uniformly continuous on $[c,d] \subseteq (a,b)$ then it is uniformly continuous on $(a,b)$? Furthermore, is it true that if $f$ is uniformly continuous on $[a,b] \ \forall a,b \in \mathbb{R}$ then $f$ is uniformly continuous on $\mathbb{R}$? 
How would I go about proving these two statements?

No: $$f:[1,2] \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$
is uniformly continuous. But 
$$f:]0,3[ \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$
is not uniformly continuous. 
If you meant for every $[c,d] \subset (a,b)$ then it is still not true:
$$\forall [c,d] \subset (0,+\infty): f:[c,d] \rightarrow \mathbb{R}: x \rightarrow \frac{1}{x}$$ is uniformly continuous.
For the second:
$$\forall [c,d] \subset (0,+\infty): g:[c,d] \rightarrow \mathbb{R}: x \rightarrow e^{x}$$ is uniformly continuous.
But
$$g:\mathbb{R} \rightarrow \mathbb{R}: x \rightarrow e^{x}$$ is not.
Let's see why this is a counterexample:
Given: $\epsilon>0$, we search: 
$$\delta>0 : \forall |x-y|<\delta \Rightarrow |e^{x}-e^{y}|<\epsilon$$
This can be rewritten:
$$|e^{x}-e^{y}|<\epsilon$$
$$e^{y}-\epsilon<e^{x}<e^{y}+\epsilon$$
Taking the $ln$ (assuming $e^{y}-\epsilon>0$):
$$ln(e^{y}-\epsilon)<x<ln(e^{y}+\epsilon)$$
Take $x = y+\frac{\delta}{2}$.
Then:
$$ln(e^{y}-\epsilon)<y+\frac{\delta}{2}<ln(e^{y}+\epsilon)$$
$$ln(e^{y}-\epsilon)-y<\frac{\delta}{2}<ln(e^{y}+\epsilon)-y$$
Taking the limit as $y \rightarrow \infty$:
$$0\leq\frac{\delta}{2}\leq0$$
Thus no such $\delta$ exists.
The problem why this doesn't work, is that $e^{x}$ increases more reapidly when $x$ gets bigger.
