This is not entirely dissimilar to the answer already posted by Chris Godsil, but I'll post this anyway, maybe it can provide slightly different angle for someone trying to understand this.
We want to show that the matrix
$$\begin{pmatrix}
x_0^0 & \cdots & x_0^n \\
\vdots & \ddots & \vdots \\
x_n^0 & \cdots & x_n^n
\end{pmatrix}$$
is invertible.
It suffices to show that the rows columns of this matrix are linearly independent.
So let us assume that $c_0v_0+c_1v_1+\dots+c_nv_n=\vec 0=(0,0,\dots,0)$, where $v_j=(x_0^j,x_1^j,\dots,x_n^j)$ is the $j$-the row column written as a vector and $c_0,\dots,c_n\in\mathbb R$.
Then we get on the $k$-th coordinate
$$c_0+c_1x_k+c_2x_k^2+\dots+c_nx_k^n=0,$$
which means that $x_k$ is a root of the polynomial $p(x)=c_0+c_1x+c_2x^2+\dots+c_nx^n$.
Now if the polynomial $p(x)$ of degree at most $n$ has $(n+1)$ different roots $x_0,x_1,\dots,x_n$, it must be the zero polynomial and we get that $c_0=c_1=\dots=c_n=0$.
This proves that the vectors $v_0,v_1,\dots,v_n$ are linearly independent. (And, in turn, we get that the given matrix is invertible.)
Hint:
Well, for one thing, $\det (A+\lambda B)$ is indeed a function of $\lambda$, a polynomial in $\lambda$ in fact. Because it is a polynomial, you can tell a lot about the roots of the function (if a polynomial satisfies $p(x)=0$ for all $x\in \mathbb R$, what can you say about $p$?)
Best Answer
Consider $$\det(A + \lambda B)$$ This will be a polynomial of finite degree in $\lambda$. Unless the polynomial is uniformly zero, there will exist only a finite set of real roots. Then any $\lambda$ which is not a root of the polynomial will give you the required invertible matrix. This does not even require $B$ to be invertible.
I'm not sure under what conditions the above will be uniformly zero (if someone knows the answer, I'd be interested to know), but I can say that when $B$ is invertible that this never happens. For we have $$\det(AB^{-1} + \lambda I) = \det(A + \lambda B)\det(B^{-1})$$ so the expressions $\det(AB^{-1} + \lambda I)$ and $\det(A + \lambda B)$ are either together zero or together non-zero. The first expression is $$-\det(-AB^{-1} - \lambda I)$$ which is the (negative) characteristic polynomial of $-AB^{-1}$, and that's always of degree $n$.