If there is $a>0$ s.t. $f|[0,a]$ and $f|[a, +\infty)$ are uniformly continuous, $f:[0, +\infty) \rightarrow \mathbb{R}$ also is Given $f:[0, +\infty) \rightarrow \mathbb{R}$, suppose that exists $a>0$ such that $f|[0,a]$ and $f|[a, +\infty)$ are uniformly continuous. Prove that $f$ is uniformly continuous.
Any hints on how to go about this? I've tried using the definition, but this is not like proving if $f$ and $g$ are u.c., then $f+g$ is as well. The best result I know is that if $f$ and $g$ are uniformly continuous and the codomain of $f$ is the same as the domain of $g$, $g \circ f$ is also uniformly continuous.
I'm not looking for a particular proof of it all, just a potential proposition that could help me see it.

The following Lemma is well-known and easy to prove (see, for instance, “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989)). Let $X$ and $Y$ be topological spaces, $\mathcal F$ be a (locally) finite cover of the space $X$ by its closed subspaces. Given a family $\{f_F:F\to Y | F\in\mathcal F\}$  of continuous functions  such that for each $F,F’\in\mathcal F$ and each point $x\in F\cap F’$ we have $f_F(x)=f_{F’}(x)$, the function $f:X\to Y$ such that $f(x)=f_F(x)$ for any point $x\in X$ and any set $x\in F\in\mathcal F$ is continuous. 
Applying Lemma to a finite closed cover $\mathcal F\{[0,a],[a,+\infty)\}$ of the space $X=[0,+\infty)$ we see that the function $f$ is continuous. In particular, the restriction $f|[0,a+1]$ of the function $f$ on a subspace $[0, a+1]$ of the space $[0,+\infty)$ is continuous. Since the set $[0, a+1]$ is compact then the function $f|[0,a+1]$ is uniformly continuous. Now let $\varepsilon>0$ be an arbitrary number. Uniform continuity of the functions $f|[0,a+1]$ and $f|[a,+\infty)$ implies that there exist numbers $0<\delta_1, \delta_2<1$ such that if $0\le x,y\le a+1$ and $|x-y|<\delta_1$ then $|f(x)-f(y)|<\varepsilon$ and 
if $a\le x,y$ and $|x-y|<\delta_2$ then $|f(x)-f(y)|<\varepsilon$. If we put $\delta=\min\{\delta_1,\delta_2\}$ then we shall have $|f(x)-f(y)|<\varepsilon$ for each $0\le x,y$ such that $|x-y|<\delta$.
