 Solution

For each $x\in X$, density of $Y$ allows a sequence $y_n\in Y$ with $d(y_n,x)<1/n$. If $f_1,f_2$ are two continuous extensions, continuity givesThus a continuous extension is unique if it exists.

Now assume $g$ is uniformly continuous. Each chosen $y_n\to x$ is a Cauchy sequence, so part (a), applied to $Y$, shows that $g(y_n)$ converges in $\mathbb{R}$. DefineThis is independent of the choice: for another $z_n\to x$ in $Y$, $d(y_n,z_n)\to 0$, and uniform continuity gives $g(y_n)-g(z_n)\to 0$. At $x\in Y$, comparison with the constant sequence $x$ proves $f(x)=g(x)$.

To prove uniform continuity of the extension, choose $\delta >0$ so that $|g(y)-g(z)|<\epsilon /2$ whenever $y,z\in Y$ and $d(y,z)<\delta$. If $d(x,x^{\prime })<\delta /3$, choose sequences $y_n\to x$ and $z_n\to x^{\prime }$ in $Y$. For large $n$, both approximation distances are less than $\delta /3$, so $d(y_n,z_n)<\delta$. Passing to the limit gives $|f(x)-f(x^{\prime })|\le \epsilon /2<\epsilon$. The same $\delta /3$ works for all $x,x^{\prime }$. Therefore the extension exists and is uniformly continuous. No completeness of $X$ is needed; completeness of the target $\mathbb{R}$ is the relevant requirement.