Define a subset $Y$ of a metric space $X$ to be totally bounded if given any $r>0$ there exists a finite open cover of $Y$ by $r$-balls of $X$,with centres in $Y$.
If $Y=X$,then we say that the metric space $X$ is totally bounded. I want to show that the following are equivalent:
$1.$ $Y$ is totally bounded in $X$.
$2.$For any $r>0$ ,there exists a finite open cover of $Y$ by balls of $X$ with radius at most $r$ and centre in $Y$.
$3.$For any $r>0$,there exists a finite open cover of $Y$ by $r$-balls of $X$.(Not necessary to be centred in $Y$).
$4.$ The metric subspace $(Y,d_Y)$ is totally bounded,where $d_Y$ is the restriction of the metric $d$ to $Y$.
Clearly $1\implies 2$.For $2\implies 1$,take the balls of radius at most $r$ and consider their centres and draw a ball of radius $r$ around each of the centres.They will also cover $Y$.So $1\iff 2$.
Clearly,$1\implies 3$.To prove that $3\implies 1$,take finitely many $r/2$ balls covering $Y$.Consider the subcollection that consist of some point of $Y$,take that point and draw $r$ ball around it.Then the $r/2$ ball with centre not necessarily in $Y$ is contained in the $r$-ball with centre in $Y$.So,the new cover by $r$-balls have centres in $Y$.So,$1\iff 3$.
$1\iff 4$ is obvius because if $Y$ is totally bounded,then for each $r>0$ there is a finite cover of $Y$ by $r$-balls of $X$.Then by intersecting each ball with $Y$,we get a finite cover or $Y$ by $r$-balls of $Y$.Conversely if $Y$ is covered by finitely many $r$-balls of $Y$,then it is also covered by corresponding $r$-balls in $X$.
We can also add another equivalent condition that $Y$ is totally bounded if for each $r>0$,$Y$ can be covered by finitely many sets with diameter $\leq r$.
So,the above statements are equivalent.Is the above proof correct?