Here's an outline of one approach:
$\ \ \ 1)$ Let $\epsilon>0$.
$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/3$ whenever $x\ge N$.
$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N]$.
$\ \ \ 4)$ Choose $ \delta>0$ so that $|f(x)-f(y)|<\epsilon/3$ whenever $|x-y|<\delta $ and $x\in[a,N]$, $y\in[a,N]$.
$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$
$\ \ \ \ \ \ \ $(consider three cases depending on the relationship between $x$, $y$, and $N$).
A slightly more elegant approach:
$\ \ \ 1)$ Let $\epsilon>0$.
$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/2$ whenever $x\ge N$.
$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N+1]$.
$\ \ \ 4)$ Choose $ 1>\delta>0$ so that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta $, $x\in[a,N+1]$, and
$\ \ \ \ \ \ $$y\in[a,N+1]$.
$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$
$\ \ \ \ \ \ \ $(consider two cases depending on whether one (or both) of $x$, $y$, exceeds $N+1$).
First, uniform continuity is about an interval, not just a point. You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$. Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement: for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.
Now what they never tell you is that $\delta$ depends on $x$. For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$. It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper. That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller, to get $|f(x) -f(c)| < \varepsilon$.
What uniform continuity means is that the $\delta$ does NOT depend on x. You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.
Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$. Steepness doesn't only mean that $f$ may go off to infinity in your interval. It could also oscillate around in some unpleasant way. Look for example at $f(x) = \sin(1/x)$ on $(0,1)$.
Best Answer
Yes, your proof is entirely correct (except you write "uniform convergence" where you mean "uniform continuity"). Well, to be extremely picky I suppose you should explain what happens when $f'(x_1) = 0$, since then you can't perform the division, but that is a trivial case.
Note that you are actually showing that $f$ is Lipschitz: we say $f: I \rightarrow \mathbb{R}$ is Lipschitz if there is a constant $C$ such that for all $x_1,x_2 \in I$, $|f(x_1)-f(x_2)| \leq C|x_1-x_2|$.
It is helpful to think in terms of the stronger conclusion, because Lipschitz functions have many other nice properties. (For instance, any Lipschitz function is absolutely continuous and thus differentiable almost everywhere. These are more advanced concepts and you may not have encountered them yet, but if you take a graduate level course on real analysis, you certainly will.)