Show that $x(p,w)=w\cdot x(p,1)$ with homothetic preferences
An indirect proof. Suppose
$$
x(p,w) = w\cdot x(p,1)
$$
does not hold. This is equivalent with stating
$$
U(x(p,w)) \neq U(w\cdot x(p,1)).
$$
(To be precise: $x(p,w)$ and $x(p,1)$ may be set valued. In this case we are talking about two elements at least one of which is not included in both sets.)

Case 1. 
$$
U(x(p,w)) &gt; U(w\cdot x(p,1))
$$
As $U$ is homothetic
$$
U(x(p,w)) =  U(w \cdot \frac{1}{w}\cdot x(p,w)) = w \cdot U(\frac{1}{w}\cdot x(p,w)).
$$
Using this we have
$$
w \cdot U(\frac{1}{w}\cdot x(p,w)) = U(x(p,w)) &gt; U(w\cdot x(p,1)) = w\cdot U(x(p,1))
$$
and thus
$$
U(\frac{1}{w}\cdot x(p,w)) &gt; U(x(p,1))
$$
However as $\frac{1}{w} \cdot x(p,w)$ is clearly an element of $B(p,1)$ this is impossible as $x(p,1)$ gives maximal utility in that budget set.

Case 2. 
$$
U(x(p,w)) &lt; U(w\cdot x(p,1))
$$
As $w \cdot x(p,1)$ is clearly an element of $B(p,w)$ this is impossible as $x(p,w)$ gives maximal utility in that budget set.

Thus we have proven that
$$
U(x(p,w)) = U(w\cdot x(p,1))
$$
which is equivalent with
$$
x(p,w) = w\cdot x(p,1).
$$