I can think of a counter-example that when $f(x)=x$, then $f(a+x)-f(x)=a$ is not bounded as $a$ is getter larger and larger. But what if I only require the bound to be independent of $x$?
Can I use extreme value theorem to say that there exists $x_0\in [x,x+a]$ such that $f(x)<f(x_0)$ on $[x,x+a]$.
In this way, we have $|f(x+a)-f(a)|<2|f(x_0)|$
But the problem is that $x_0$ may implicitly depends on $x$, so I am confused.
On
Let's be concrete here, and let me show that $g(x):=f(1+x)-f(x)$ is a bounded function in $x$.
Since $f$ is uniformly continuous, there is a positive real number $\delta$ such that for every pair of real numbers $x$ and $y$ that satisfy the inequality $|y-x|<\delta$, we may assert that $$|f(y)-f(x)|<1\,.$$
Now choose a natural number $n$ so that $0<1/n<\delta$. Then, regardless of $x$, we can say that $$|f(x+1/n)-f(x)|<1\,.$$
Which in turn implies that, for any $x$, we can say \begin{align*}|f(x+2/n)-f(x)|&=|f(x+2/n)-f(x+1/n)+f(x+1/n)-f(x)|\\&\leq |f(x+2/n)-f(x+1/n)|+|f(x+1/n)-f(x)|\\&<1+1\\&=2 \end{align*} This reasoning could be evidently expanded. By induction, we can show $$|f(x+k/n)-f(x)|<k$$ for any natural number $k$. In particular, $$|f(x+1)-f(x)|<n\,.$$
Thus, $g(x)$ is a bounded function: $|g(x)|<n$ for all $x$. But there's nothing special about the number $1$. Replace it with $a$ or $|a|$ where appropriate to get an argument.
By definition of uniform continuity, there is $\epsilon>0$ such that $|f(x)-f(y)|<1$ whenever $|x-y|<\epsilon$. There is $N\in\Bbb N$ with $|a/N|<\epsilon$.
Now prove that $|f(x+a)-f(x)|<N$ for all $x$.