True or False: If $f \colon [a, \infty) \mathbb{R}$ is bounded and continuous, then $f$ is uniformly continuous on $[a, \infty)$ True or False: If $f \colon [a, \infty) \rightarrow \mathbb{R}$ is bounded and continuous, then $f$ is uniformly continuous on $[a, \infty)$. 
If true, supply a proof. If false, provide a counter example.
So my thinking is that it is true. I couldn't think of a counter example for this so that's why I believe it is true. What I am having trouble with is showing that it is true. I know the definition for $f$ to be continuous is: $\forall \epsilon >0, \exists \delta > 0,$ such that if $x \in [a, \infty$) and $|x-c|< \delta$ then $|f(x)-f(c)|< \delta$. 
I know the definition for $f$ to be uniformly continious is $\forall \epsilon >0, \exists \delta > 0,$ such that if $x,y \in [a, \infty$) nd $|x-y|< \delta$
then $|f(x)-f(y)|< \delta$. 
I was sort of thinking about "combining" these two definitions, however the $\delta$ in the first definition is dependent on $c$ and the $\delta$ in the second definition CANT depend on $c$ (in this case $y$).
Is this true? can someone hep me prove this 

Consider $f(x)=\sin x^2$ .
Then $ x_n=\sqrt{\dfrac{(n+1)\pi}{2}}$ and $y_n=\sqrt{\dfrac{n\pi}{2}}$
Choose $\delta$ such that $|x_n-y_n|<\delta $ for $n $ large but we have $|\sin(x_n)-\sin (y_n)|=1$
