Show for $f:A \to Y$ uniformly continuous exists a unique extension to $\overline{A}$, which is uniformly continuous

(Your proof above should explicitly show that $g$ is independent of the sequence used to define it. This is the key point of the proof.)

Let $\epsilon>0$, then you have some $\delta>0$ such that if $d(x,y) < \delta$, then $d(f(x),f(y)) < {1 \over 2}\epsilon$.

Pick $x,y \in \overline{A}$ such that $d(x,y) < \delta$, and let $x_n,y_n$ be sequences in $A$ such that $x_n \to x,y_n \to y$. By construction above, $g(x) = \lim_n f(x_n)$ and similarly for $g(y)$.

For sufficiently large $n$, we have $d(x_n,y_n) < \delta$, and so $d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$.

Taking limits we have $d(g(x),g(y)) \le {1 \over 2}\epsilon < \epsilon$.