Question about proof of the tube lemma for metric spaces Tube lemma:
Let $M$ be a metric space and $K$ a compact metric space. Let $a\in M$, $a\times K\subset V\subset M\times K$, that is, suppose there is an open set $V$ between $a\times K$ and $M\times K$. Tube lemma says that then there exists an open $U$ such that $a\times K\subset U\times K \subset V\subset M\times K$ 
The proof follows like this and use the image below:
Since $a\times K$ is homeomorph to $K$, it is compact. Therefore, there exists $r>0$ (WHY?)such that: $z\in (M\times K)-V, t\in K\implies d(z,(a,t))\ge r$ (WHY?). Pick $U = B(a, t)$ in $M$. The, $(x,t)\in U\times K\implies d(x,a)<r\implies (x,t)\in V$, so $U\times K\in V$.
The first why: how does compact relates to the existence of such $r>0$? Is it because we can find a finite open subcover? Why would this $r$ be so great that it can escape $V$? Is it because the definition of compacts for covers must work for all covers?
For the second why: why does he pick a point outside V??? Shouldn't it be inside? That makes no sense! Could someone explain this part better to me? 

 A: Assuming that you’ve reproduced it accurately, that proof makes life difficult by using $d$ both for the metric on $M\times K$ and for the metric on $M$. In fact the metrics just get in the way: the result is true for an arbitrary space $M$ and an arbitrary compact space $K$, and I think that the proof is actually a bit clearer in that setting, so let me give that first.
For each $t$ in $K$ there are open nbhds $B_t$ of $a$ in $M$ and $N_t$ of $t$ in $K$ such that $B_t\times N_t\subseteq V$; this is just from the definition of the product topology on $M\times K$. $\{N_t:t\in K\}$ is then an open cover of $K$, so there is some finite $\{t_1,\ldots,t_n\}\subseteq K$ such that
$$K=\bigcup_{i=1}^nN_T{t_i}\;.$$
Let $$U=\bigcap_{i=1}^nB_{t_i}\;;$$ $B$ is an open nbhd of $a$, and 
$$\{a\}\times K\subseteq\bigcup_{i=1}^n(U\times N_{t_i})\subseteq\bigcup_{i=1}^n(B_{t_i}\times N_{t_i})\subseteq V\;.$$
Finally,
$$\bigcup_{i=1}^n(U\times N_{t_i})=U\times K\;,$$
so we’re done: $U$ has the required properties.
In the metric case the details missing from the argument that you reproduced — the ones justifying the existence of $r$, for instance — depend a bit on exactly how the metric on the product $M\times K$ is defined in terms of metrics on $M$ and $K$; without knowing that, I can’t fill them in explicitly. 
A: For each $t\in K$ let $M_t$ be open in $M$  and let $K_t$ be  open in $K $ such that $$ (a,t)\in M_t\times K_t\subset V.$$ Since $\{K_t:t\in K\}$ is an open cover of the compact space $K,$ we can take a finite $F\subset K$ such that $\cup \{M_t :t\in F\}= K .$
Now let $U=\cap_{t\in F}M_t.$ Then $a\in U\subset M_t$ and $M_t\times K_t\subset V $ for each $t\in F, $ so we have  $$\{a\}\times K\subset U\times K= U\times (\cup_{t\in F}K_t)=$$ $$=\cup_{t\in F}(U\times K_t)\subset \cup_{t\in F}(M_t\times K_t)\subset V.$$  
Remark: This is the same proof as by Brian M. Scott, which I did not study before posting this.
