Let $X$ be a metric space and $q \in X$. Suppose that sequence $\left\{ p_k \right\}$ in $X$ converges to a point $p$ in $X$. Then, $\left\{ d(p_k,q) \right\}$ converges to $d(p,q)$.
\begin{align}
& \therefore \forall \varepsilon \exists N \forall k (k \ge N
\implies d(p_k,p) < \varepsilon) \label{eq2} \\
& \abs{d(p_k,q) - d(p,q)} < \varepsilon \\
& \iff d(p_k,q) - d(p,q) <
\varepsilon \land d(p,q) - d(p_k,q) < \varepsilon \notag \\
& d(p_k,q) - d(p,q) \le d(p,p_k) \iff d(p_k,q) \le
d(p,q) + d(p,p_k) \label{eq4} \\
& d(p,q) - d(p_k,q) \le d(p,p_k) \iff d(p,q) \le
d(p_k,q) + d(p,p_k) \label{eq5}
\end{align}
\eqref{eq4}, and \eqref{eq5} follows from the Triangular Inequality. Apply \eqref{eq2} to \eqref{eq4} and \eqref{eq5} to finish the proof. Q.E.D.
Remark: Quantifications similar to \eqref{eq2} can be found in Wikipedia’s entry for uniform continuity.