real analysis - A proof about uniformly continuous function on an open interval and its extension

I try to prove:

If $g$ is a continuous function on $(a,b)$ then $g$ is uniformly continuous on $(a,b)$ if and only if it is possible to define values $g(a)$ and $g(b)$ at the endpoints so that the extended function $g$ is continuous on $[a,b]$.

Please can you check my proof?

If there are $g(a)$ and $g(b)$ such that $g:[a,b]\to \mathbb R$ is continuous then because $g:[a,b]\to \mathbb R$ is uniformly continuous, so is $g:(a,b)\to \mathbb R$.

Let $g$ be uniformly continuous on $(a,b)$ and let $x_n \to a$ and $y_n \to b$. Because $x_n$ and $y_n$ are Cauchy and $g$ is uniformly continuous, $g(x_n)$ and $g(y_n)$ are Cauchy. Define $g(a) = \lim g(x_n)$ and $g(b) = \lim g(y_n)$. Now this extended function $g$ is continuous at $a$ and $b$. It is enough to show it for one of the two points because the proof is similar. Let $\varepsilon >0$. The goal is to find $\delta$ such that $|x-a|<\delta$ implies that $|g(x) -g(a)|<\varepsilon$. By definition, $g(x_n) \to g(a)$ hence there is $N$ such that for $n>N$ it holds that $|g(x_n) - g(a)| < \varepsilon$. Now let $\delta = |a-x_{N+1}|$.

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I now know why my proof is false, thanks to the answer but I can't understand one part of the answer. The bounty is for extension of answer not for more answers.

Answer

To show the extended function $g$ is continuous at $a$ let $\varepsilon > 0$.

Note that $|x-x_n|\le |x-a| + |a-x_n|$. Let $\delta$ be such that $|x-y|<\delta \implies |g(x)-g(y)|<{\varepsilon \over 2}$.

Let $N$ be such that $n>N$ implies that $|a-x_n|<{\delta \over 2}$ and let $N'$ be such that $n>N'$ implies $|g(x_n)-g(a)|<{\varepsilon \over 2}$.

Now if $|x-a|<{\delta \over 2}$ and $n>\max(N,N')$ then $|x-x_n|\le |x-a| + |a-x_n|<\delta$ and hence

$$|g(x)-g(a)| \le |g(x)-g(x_n)| + |g(x_n) - g(a)| < \varepsilon$$