I'm trying to prove that if $X_1,X_2,X_3 \in \mathbb{R}$ are pairwise independent, then $X_1$ and $(X_2,X_3) \in \mathbb{R}^2$ are independent.
My attempt: We know
$$ f_{X_1,X_2}(x_1,x_2) = f_{X_1}(x_2)f_{X_2}(x_2) $$ $$ f_{X_1,X_3}(x_1,x_3) = f_{X_1}(x_3)f_{X_2}(x_3) $$ $$ f_{X_2,X_3}(x_2,x_3) = f_{X_2}(x_2)f_{X_3}(x_3) $$
Now we need to show
$$ f_{X_1,(X_2,X_3)}(x_1,(x_2,x_3)) = f_{X_1}(x_1)f_{X_2,X_3}(x_2,x_3) $$
We know
$$f_{X_1}(x_1)f_{X_2,X_3}(x_2,x_3) = f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(x_3)$$
I'm stuck here.
What you are trying to prove is not true.
Throw two numbered coins ($1$ and $2$) and let $X_i$ take value $1$ if coin $i$ shows heads and value $0$ otherwise. Further let $X_3$ take value $1$ if $X_1=X_2$ and value $0$ otherwise.
Then pairwise independence for $X_1,X_2,X_3$ is not difficult to verify, but:$$P(X_1=1\wedge (X_2,X_3)=(1,1))=P(X_1=1=X_2)=\frac14$$ while:$$P(X_1=1)P((X_2,X_3)=(1,1))=P(X_1=1)P(X_1=1=X_2)=\frac12\frac14=\frac18$$
So $X_1$ and $(X_2,X_3)$ are not independent.