I'm reading the proof in Rudin and I came into this:
Let $E$ be a dense subset of a metric space $X$and let $f$ be a uniformly continuous real function defined on $E$.
... $g$ is an extension of $f$. Since $f$ is uniformly continuous on $E$ and $g(x) = f(x)$ for $x \in E$, then $g$ is uniformly continuous on $E$.To show that g is uniformly continuous on X, we need to show that for an arbitrary point $y \in X \cap E^c$ and any $\epsilon>0$ there exists $\delta>0$ such that $d_Y(g(x), g(y)) < \epsilon$ whenever $x \in X\cap E^c$ and $d_X(x,y)<\delta$. Recall that since $y \in X \cap E^c$ there is a sequence of points $\{p_n\}$ in $E$ which converges to $y$ and that $\{g(p_n)\}$is a Cauchy sequence in Y which converges to what we call $g(y)$.
I don't get this part. Why is this true? Can smb explain it? Thanks.
That sequence $(p_n)_{n\in\mathbb N}$ exists because $E$ is dense in $X$. Since $g|_E$ is uniformly continuous and $(p_n)_{n\in\mathbb N}$ is a Cauchy sequence (since it converges), $\bigl(g(p_n)\bigr)_{n\in\mathbb N}$ is also a Cauchy sequence. You told us nothing about the space $Y$, but my feeling is that it is complete. Am I right? If so, then the sequence $\bigl(g(p_n)\bigr)_{n\in\mathbb N}$ converges.