Product of metric spaces 
Let $(M,d)$ and $(N,p)$ be metric spaces. Consider the space $M \times N$  endowed with the metric $D=((x,y),(x',y'))=\max\{d(x,x'),p(y,y')\}$ for $(x,y),(x',y') \in M \times N$.
Let $A \subseteq M$ and $B \subseteq N$ be nonempty. Prove or disprove:
a) $P: M\times N \to M$ given by $P(x,y)=x$ is continuous and an open mapping
b) The set $A \times B$ is totally bounded in $M\times N$ if and only if $A$ and $B$ are totally bounded in $M$ and $N$, respectively
c) The set $A\times B$ is connected in $M\times N$ if and only if $A$ and $B$ are connected in $M$ and $N$, respectively

I feel like the answers are true, true, false. I have no clue how to show any of them though. I'm not sure how to work with the endowed metric.
 A: We use the notation $B_\rho(a, r)$ to denote the open ball centered at $a$ with radius $r$ measured by the metric $\rho$.
a) To show that $P$ is a continuous open mapping, show that for any $\varepsilon > 0$ and fixed $(x_0, y_0) \in M \times N$, we have $P\big( B_D((x_0,y_0), \varepsilon) \big) = B_d(x_0, \varepsilon)$. [Why does this suffice? How do we show this?]
b) We rename the map $P$ in part (a) with $\pi_x$. By symmetry and part (a), the map $\pi_y: M \times N \to N$ given by $\pi_y(x,y) = y$ is open and continuous. If $A$ is totally bounded by balls $A_1, \dots, A_m \subseteq M$ and $B$ is totally bounded by balls $B_1, \dots, B_n \subseteq N$, then how can we make balls totally bounding $A \times B \subseteq M \times N$? [Hint: $\{A_i \times B_j : 1 \leq i \leq m \mbox{ and } 1 \leq j \leq n\}$.] Conversely, given balls, (say $U_1, \dots, U_l$) in $M\times N$ totally bounding $A \times B$, how to we get balls in $M$ totally bounding $A$ and balls in $N$ totally bounding $B$? [Hint: $\{\pi_x U_i\}_{i=1}^l$ and $\{\pi_y U_i\}_{i=1}^l$.] The hints give you the "answers" but you must fill in the details. (Ask specific questions if this is still troubling you.)
c) What's your intuition on why it's false? Can you give a counter-example?
