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
$\ \ \ \ \ \ $$y\in[a,N+1]$.
$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$
$\ \ \ \ \ \ \ $(consider two cases depending on whether one (or both) of $x$, $y$, exceeds $N+1$).
Yes, your proof is entirely correct (except you write "uniform convergence" where you mean "uniform continuity"). Well, to be extremely picky I suppose you should explain what happens when $f'(x_1) = 0$, since then you can't perform the division, but that is a trivial case.
Note that you are actually showing that $f$ is Lipschitz: we say $f: I \rightarrow \mathbb{R}$ is Lipschitz if there is a constant $C$ such that for all $x_1,x_2 \in I$, $|f(x_1)-f(x_2)| \leq C|x_1-x_2|$.
It is helpful to think in terms of the stronger conclusion, because Lipschitz functions have many other nice properties. (For instance, any Lipschitz function is absolutely continuous and thus differentiable almost everywhere. These are more advanced concepts and you may not have encountered them yet, but if you take a graduate level course on real analysis, you certainly will.)
Best Answer
Let $M>0$ be an upper bound for $|f(x)|$, and $\epsilon > 0$ be given. From the given condition it follows that there is a $\delta > 0$ such that $$ |f(x+h)-2f(x)+f(x-h)| < \epsilon $$ for all $x \in \Bbb R$ and all $h \in (0, \delta)$. We will now show that $$ \tag{*} |f(x) - f(y) | \le 2 \sqrt{M \epsilon} $$ for all $x, y\in \Bbb R$ with $|x-y| < \delta$, which implies that $f$ is uniformly continuous.
In order to prove $(*)$ we use the following result from Prove or disprove $\lim\limits_{n \to \infty}\Delta x_n=0.$ :
For fixed $x, y \in \Bbb R$ with $0 < |x-y| < \delta$ we can apply this result to the sequence $$ f_n = f(x + n(y-x)) $$ which satisfies $|f_n| \le M$ and $|\Delta^2 f_n| \le \epsilon$. It follows that $$ |f(x) -f(y) | = |\Delta f_0| \le 2 \sqrt{M \epsilon} $$ and this completes the proof.