problem about uniform continuity Problem:

"Show that if $f:\left[a,+\infty\right] \rightarrow \mathbb R$ is continuous and if $\lim_{x \rightarrow + \infty} f(x)$ exists, then $f$ is uniformly continuous".

I have a question about this problem.
My partial solution:
$\lim_{x\rightarrow +\infty}f(x) = L \Rightarrow \forall \varepsilon>0, \exists M>0$ such that $x \geq M \Rightarrow  |f(x)-L|<\frac{\varepsilon}{2}$.
Given $x_1$,$x_2$ $\in$$[M, \infty)$$\Rightarrow |f(x_1)-L|<\frac{\varepsilon}{2}$ and $|f(x_2)-L|<\frac{\varepsilon}{2}$ $\Rightarrow |f(x_1)-L|<\frac{\varepsilon}{2}$ and $|L-f(x_2)|<\frac{\varepsilon}{2}$ $\Rightarrow |f(x_1)-L+L-f(x_2)|\leq |f(x_1)-L|+|L-f(x_2)|< \varepsilon$ $\Rightarrow |f(x_1) - f(x_2)|<\varepsilon$ $\Rightarrow$ $f$ is uniformly continuous on $[M, +\infty)$.
How can I extend this result towards $f$ uniformly continuous on $[a, +\infty]$?
 A: 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 $N+1$).
