Exterior Covering Number of $\epsilon /2$ is greater of equal than Covering Number of $\epsilon$ Definition $(\epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K \subset T$ and let $\epsilon >0$, A subset $N \subset K $ is called $\epsilon -Net$ of $K$ if every point in $K$ is within a distance  $\epsilon$ if every point in K is within a distance $\epsilon$ of some point of $N$,i.e.
$$\forall x \in K  \ \ \  \exists x_0 \in N \ : d (x,x_0) \leq \ \epsilon$$
Definition(Covering  Number): For  metric space  $(T,d)$ The covering number of $K \subset T$  respect to a given $\epsilon \geq 0$ ,denotes as $N(K,d,\epsilon )$, is the smallest possible cardinarity an  $\epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers  in $K$ and radii $\epsilon$ whose union covers $K$ 
Definition(Exterior Covering  Number): For  metric space  $(T,d)$ The exterior covering number of $K \subset T$  respect to a given $\epsilon \geq 0$ ,denotes as $N^{ext}(K,d,\epsilon )$,  is the smallest number of closed balls with centers not necessary in  $K$ and radii $\epsilon$ whose union covers $K$ 
then  I was  asked to prove that:
$$N(K,d,\epsilon) \leq N^{ext} (K,d,\epsilon /2) $$
how to see that ?
here is my attempt: since each $\epsilon -ball$ in $N(K,d, \epsilon) $ should intersect at least one  $\epsilon /2 -ball$ in $N^{ext}(N,d,\epsilon/2)$ ,thus I am trying to show that by contradiction: for each two  distinct $\epsilon -ball$ in $N(K,d, \epsilon)$ ,the $\epsilon /2 -balls$ they intersect must contain a distinct one.

Suppose that $$\bar B(x_1, \epsilon / 2), \ \dots, \ \bar B(x_{N^{\rm ext}}, \epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i \in \{ 1, \dots, N^{\rm ext}\}$, there exists a $k_i \in K$ that is contained in $\bar B(x_i, \epsilon / 2)$. (Otherwise $\bar B(x_i, \epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering). 
By the triangle inequality,
$$ \bar B(x_i , \epsilon / 2) \subseteq \bar B(k_i, \epsilon), $$
for each $i$, which means that
$$\bar B(k_1, \epsilon ), \ \dots, \ \bar B(k_{N^{\rm ext}}, \epsilon)$$
is an internal covering of size $N^{\rm ext}$. Hence the smallest internal covering has size at most $N^{\rm ext}$.
