Proving if f has a continuous extension to [a,b] then f is uniformly continuous on (a,b). I want to show that if a function $f:(a,b) \rightarrow \mathbb{R}$ has a continuous extension to [a,b], then f is uniformly continuous. Also, assume that $(X, d_x)$ and $(Y, d_y)$ are nonempty metric spaces and all subsets of $\mathbb{R}^k$ are given the Euclidean metric. 
I know the proof is trivial, but I'm worried about my notation as I don't think I quite understand continuous extension correctly. I understand that for $E \subset F \subset X$, the function $g: F \rightarrow Y$ is an extension to f if $g(x)=f(x) \forall x \in E$. 
So, we have $(a,b) \subset [a,b] \subset \mathbb{R}$ and $f: (a,b) \rightarrow \mathbb{R}$. So if f has a continuous extension to a,b, then (f(x)) = f(y)) $\forall x \in (a,b), y \in [a,b]$. And so this then implies that d(f(x), f(y)) = 0, and so $\forall \epsilon > 0, \exists \delta > 0$ s.t. $d(x, y) < \delta$ implies $d(f(x), f(y)) < \epsilon$. 
Help and hints would be much appreciated!

Let's get intuitive here :
Saying "$f:(a,b)\to\mathbb{R}$ has a continuous extension to $[a,b]$" is to say "we can define $f(a),f(b)$ so that f is continuous on $[a,b]$"
One example of this would be $f:(0,+\infty)\to\mathbb{R}$ with $f(x) = {\sin x\over x}$. Obviously you can't compute $f(0)$ but we can say arbitrarily $f(0)=1$ and it would work.
This work only with 'nice' functions that doesn't behave wierdly or explodes to $\infty$ at its bound values.
Now back to your example.If $f$ can be continuously extended to $[a,b]$ and is continuous on $(a,b)$(you didn't say it in your post but we'll assume we have this condition) then $f$ is uniformly continuous on $[a,b]$ (due to $[a,b]$ being a close interval).
And if $f$ is uniformly continuous on $[a,b]$ it is indeed uniformly continuous on $(a,b)\subset[a,b]$
