real analysis - Prove that if $g(x)$ is uniformly continuous in $(a,b]$ and $[b,c)$ then is uniformly continuous in $(a,c)$

I open this question to check my own proof and to ask a related question.

My proof: if $g(x)$ is uniformly continuous in $(a,b]$ and $[b,c)$ then

$$\forall\varepsilon_1>0,\exists\delta_1>0,\forall x\in (a,b]:|x-b|<\delta_1\implies|g(x)-g(b)|<\varepsilon_1$$
$$\forall\varepsilon_2>0,\exists\delta_2>0,\forall y\in [b,c):|y-b|<\delta_2\implies|g(y)-g(b)|<\varepsilon_2$$

If we call $\varepsilon_0=\varepsilon_1+\varepsilon_2$ and $\delta_0=\delta_1+\delta_2$ and due triangle inequality

$$|x-y|\le|x-b|+|y-b|\\|g(x)-g(y)|\le|g(x)-g(b)|+|g(y)-g(b)|$$

Then we have the case that

$$\forall\varepsilon_0>0,\exists\delta_0>0,\forall x\in (a,b]\land\forall y\in [b,c):|x-y|<\delta_0\implies|g(x)-g(y)|<\varepsilon_0\tag{1}$$

And by definition of $g(x)$ we have too that

$$\forall\varepsilon_0>0,\exists\delta_a>0,\forall x,y\in (a,b]:|x-y|<\delta_a\implies|g(x)-g(y)|<\varepsilon_0\tag{2}$$
$$\forall\varepsilon_0>0,\exists\delta_b>0,\forall x,y\in [b,c):|x-y|<\delta_b\implies|g(x)-g(y)|<\varepsilon_0\tag{3}$$

Cause $(1)$, $(2)$ and $(3)$ if we took $\delta_{\omega}=\min\{\delta_0,\delta_a,\delta_b\}$ then we can finally write

$$\forall\varepsilon_0>0,\exists\delta_{\omega}>0,\forall x,y\in (a,c):|x-y|<\delta_{\omega}\implies|g(x)-g(y)|<\varepsilon_0$$

Two questions:

1. Is my proof right? I think is right but Im not completely sure.


2. Can you tell me some different $\delta{-}\varepsilon$ proof for the same problem?

Thank you in advance.

Answer

Let $\varepsilon>0$. By uniform continuity on $(a,b]$ and $[b,c)$, there is $\delta_1>0$ s.t. $$\forall x,y\in (a,b],\ |x-y|<\delta_1\implies |g(x)-g(y)|<\frac{\varepsilon}{2}$$
and there is $\delta_2>0$ s.t. $$\forall x,y\in [b,c),\ |x-y|<\delta_2\implies |g(x)-g(y)|<\frac{\varepsilon}{2}.$$
Let $\delta=\min\{\delta_1,\delta_2\}$ and $x\leq b\leq y$ s.t. $|x-y|\leq \delta$. Then,
$$|g(x)-g(y)|\leq |g(x)-g(b)|+|g(y)-g(b)|\leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

As you see, $\varepsilon$ is fixed at the beginning. Since it's unspecified, we have the result for all $\varepsilon>0$, and this prove the claim.