Let $G=GL_2(F)$, where $F$ is a non-Archimedean local field, for example, you can take the finite extension of $\mathbb{Q}_p$ or $\mathbb{C}_p$.
We devote $Z(G)[G,G]$ is the subgroup generated by the center $Z(G)$ of $G$ and the derived subgroup $[G,G]$. So the question is why $Z(G)[G,G]$ is of finite index in $G$?
Thanks!
It's well known that in the situation at hand, the centre of $G:= GL_n(K)$ consists of the scalar matrices $\lbrace diag(a,...,a) = a \cdot I_n : a \in K^\times \rbrace$, and its derived subgroup is $SL_n(K)$ (the last assertion is true for any field except if $n=2$ and $K = \mathbb F_2$ or $\mathbb F_3$).
Now consider the group homomorphism $\det: G \twoheadrightarrow K^\times$ whose kernel is $[G,G] = SL_n(K)$ so that we have an isomorphism
$$G/[G,G] \simeq K^\times.$$
Note that an inverse of this map is given by $x \mapsto diag(x,1,...,1) \cdot [G,G]$, and since $diag(a,...,a) \cdot SL_n(K) = diag(a^n,1,...,1) \cdot SL_n(K)$, this isomorphism identifies $Z(G)$ with $(K^\times)^n$, so we get an induced isomorphism
$$G/(Z(G) \cdot [G,G]) \simeq K^\times/(K^\times)^n.$$
After thus translating to a question about the structure of the multiplicative group of $K$, can you finish this with specific knowledge about local fields? Note however (cf. YCor's comment) that the statement will not be true if the characteristic of $K$ divides $n$ (in your case $GL_2$, iff $K \simeq F_{2^r}((t))$).