Let $(X,d)$ be a metric space and $T$ the induced topology on $X$ by $d$.
Prove that the function $d$ from the product space $(X,T) \times (X,T)$ into $\Bbb R$ is continuous.
My question is what does it mean that "the function d" is continuous?
Does it mean that [using the inverse image of an open set in $\Bbb R$ (call it $V$) is an open set in $(X,T) \times (X,T)$] $$f^{-1}(V) = \{<x_1,x_2> \in (X,T) \times (X,T) : d(<x_a, x_{b}>, <x_1, x_2>) \in V\}$$ for some fixed $<x_a, x_b> \in (X,T) \times (X,T)$?
First of all the notation $(X,T)\times (X,T)$ is somehow unusual. You could say that you take $X\times X$ with the product topology.
Now that this is out of the way, how do open sets of $(X,T)$ look like? These are unions of "balls" of the form: $B(x,r)=\{y\in X\vert d(x,y)<r\}$, where $x \in X$.
The function $d:X\times X \to \mathbb{R}$ is just that. A function. So let $V$ be a open set in $\mathbb{R}$, we can check for continuity by looking at the preimage. Now if $d^{-1}(V)$ is empty we are done since the empty set is open. So Assume $d^{-1}(V)$ is not empty. Now we can form the following open sets, for $(x,y) \in d^{-1}(V)$ let $r=d(x,y)$ and since $V$ is open there is a $\varepsilon$ s.t. $(r-2\varepsilon,r+2\varepsilon)\subset V$ define:
$$ B_x=\{z \in X\vert r-\varepsilon<d(x,z)<r+\varepsilon \}\\ B_y=\{z \in X\vert r-\varepsilon<d(z,y)<r+\varepsilon \}. $$
Firstly note that these are open sets, since it is just $B(x,r+\varepsilon)\setminus \overline{B(x,r-\varepsilon)}$. Now $U:=B_x\times B_y$ is a open set in $X\times X$ and you can check that $U\subset d^{-1}(V)$ (You should do this). Since $(x,y)\in d^{-1}(V)$ was arbitrary and we found a open neighborhood of $(x,y)$ in $d^{-1}(V)$ we have proved that $d^{-1}(V)$is open.