Example of metric continuous with respect to another metric but generating different topology Take, say, the standard 2-sphere $S^2$.  Equip it with some metric $d$; this metric will generate a topology that may or may not coincide with the standard Euclidean topology.  In the case it does, then the functions $d_x := d(\cdot, x)$ will be continuous with respect to the Euclidean topology.  My question is: what is an example when the converse does not hold?  Specifically, give an example of a metric space $(S^2,d)$ where all functions $d_x := d(\cdot, x)$ are continuous with respect to the Euclidean metric but generating a different topology.  (Note in this case that the Euclidean topology is finer than that generated by $d$.) 
 A: In this case no such solution exists.  Let $d_E$ be the Euclidean distance.  As you note, if $d(x, \cdot)$ is $d_E$-continuous for each $x$ then $(S^2, d)$ will be a coarser topology than $(S^2, d_E)$.  So the identity map of $S^2$, considered as a map from $(S^2, d_E)$ to $(S^2, d)$ is a continuous bijection from a compact space to a Hausdorff space; any such map is a homeomorphism.
To see it more directly, it suffices to show any $d_E$-closed set $K$ is $d$-closed.  Since $S^2$ is $d_E$-compact, so is $K$.  If $\{U_\alpha\}$ is any cover of $K$ by $d$-open sets, then the $U_\alpha$ are also $d_E$-open and so by compactness there is a finite subcover $\{U_{\alpha_i}\}$.  Thus $K$ is $d$-compact and hence $d$-closed.
For a non-compact example, let $X = C([0,1])$ be the set of all continuous real-valued functions on $[0,1]$, equipped with the uniform metric $d_0(f,g) = \sup_{x \in [0,1]} |f(x)-g(x)|$.  Let $d(f,g) = \int_0^1 |f(x)-g(x)|\,dx$ be the $L^1$ metric.  If $f, g, h \in C([0,1])$, we have
$$\begin{align*} |d(f,g) - d(f,h)| &\le d(g,h) && \text{ (triangle inequality)} \\ &= \int_0^1 |g(x) - h(x)|\,dx \\ &\le \sup_{x \in [0,1]} |g(x) - h(x)| \\ &= d_0(g,h) \end{align*}$$
so $d(f, \cdot)$ is continuous (even Lipschitz).  But they do not generate the same topology.  If we let $f_n(x) = x^n$ and $f(x) = 0$ then it is an easy computation to see that $d_0(f_n, f) = 1$ while $d(f_n, f) \to 0$.  So $f_n \to f$ in the $d$ metric but not in $d_0$.
