Show that $d$ is continuous on $M \times M$, where $M \times M$ is supplied with "the" product metric (referring to any one of $d_1$, $d_2$, or $d_\infty$). This says that $d$ is jointly continuous, that is, continuous as a function of two variables. [Hint: if $x_n \to x$ and $y_n \to y$, show that $d(x_n, y_n) \to d(x, y)$]
I don't completely understand this question. Using the hint provided by Carothers, I came up with the following proof:
Choose a point $q = (x, y)\in M \times M$ along with convergent sequences $x_n, y_n \in M$ where $x_n \to x$ and $y_n \to y$. Setting $N$ sufficiently high so that $|x_n - x|$ and $|y_n - y|$ are less than $ \epsilon/2$ for all $n\geq N$, we find that the distance between $q$ and $q_n = (x_n, y_n)$ is $$d^p(q_n, q) = (|x_n - x|^p + |y_n - y|^p)^{1/p} < \epsilon \text{ for all }n \geq N\\ \text{or equivalently, that }d(x_n, y_n) \to d(x, y)$$
Where I'm having trouble is in understanding how this maps onto the definition of continuity using pre-images, i.e., $B_\delta^d(x) \subset f^{-1}(B_\epsilon^\rho(f(x))$.
If the distance function is the map $f: (M\times M, d) \to ([0, \infty), d)$, are we simply defining $f(q^i) = d(q^i, q)$ and then saying that points in the $\delta < \epsilon$ open neighborhood of $q$ in $M\times M$ are in the preimage of the open interval $[0, \epsilon)$ of $[0, \infty)?$
Edit: Just to clarify, I believe I understand the relationship between the sequential and topological definitions of continuity, but was confused over whether the question is asking me to establish the continuity of $d^p$ using the metric itself, i.e., whether I'm being asked to show that $d^p(B_\delta^{d_p}(q)) \subset B_\epsilon^{d_p}(d^p(q))$.