$f$ is uniformly continuous on $(a, b]$ implies $\lim\limits_{x\to a^+} f(x)$ exists and finite

Question

I need to prove:

$f$ is uniformly continuous on $(a, b]$ implies $\lim\limits_{x\to a^+} f(x)$ exists and finite

Now, I already have a sketch for the proof:

1. Let $\{x_n\}$, a sequence such that $a<x_n\rightarrow a$ and show (Using the fact that $f$ is uniformly continuous) that $\{f(x_n)\}$ Cauchy therefore it is convergent.
2. Given $\{x_n\}, \{y_n\}$ two sequences which convergent to $a$ (from the right), construct $\{z_n\}$ such that $z_{2k}=x_{2k}$ and $z_{2k+1}=y_{2k+1} \ \forall k\in\mathbb N$. Show that $\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(y_n)$
3. Show $f(z_n)$ converges as well to the same $L$ as $f(x_n)$ and $f(y_n)$.
4. Conclude: $\lim_{x\to a^+}f(x)$ exists.

My Question:
Isn't it sufficient to stop at (1)? $\{x_n\}$ was chosen arbitrarily, So why do I need to take another two arbitrary sequences ($\{x_n\}, \{y_n\}$), construct from them a third sequence ($\{z_n\}$) and only then conclude $\lim_{x\to a^+}f(x)$ exists?

I'm missing the point of all that. I'd be glad for an explanation

Answer 1

Imagine for two Cauchy sequence $(x_n)_n$ and $(y_n)_n$, $f(x_n)\to L$ and $f(y_n)\to L'$ with $L\neq L'$ in this case $f$ has no limit in $a$.