To establish that $U_k \cap U_{k+1} \subset W_{n-k+1}$:

Consider the polynomial $P(t)=\prod (t-x_i)$. The elementary symmetric polynomials $\sigma_i$ are its coefficients, up to the sign.

Suppose that $\sigma_k=\sigma_{k+1}=0$. This means that $0$ is a multiple root of the derivative of order $(n-k-1)$ of $P$. Now you can conclude that $0$ is a root of $P$ of multiplicity at least $n-k+1$ (that is: at least $n-k+1$ of the numbers $x_i$ are zero) by means of the following facts:

If a non-constant polynomial $Q$ has all its roots real, then so does its derivative $Q'$. Then the roots of $Q'$ are the multiple roots of $Q$, plus one simple root between each pair of consecutive distinct roots of $Q$. In particular, if $x$ is a multiple root of $Q'$ then it is also necessarily a root of $Q$.

To establish that $U_k \cap U_{k+1} \subset W_{n-k+1}$:

Consider the polynomial $P(t)=\prod (t-x_i)$. The elementary symmetric polynomials $\sigma_i$ are its coefficients, up to the sign.

Suppose that $\sigma_k=\sigma_{k+1}=0$. This means that $0$ is a multiple root of the derivative of order $(n-k-1)$ of $P$. Now you can conclude that $0$ is a root of $P$ of multiplicity at least $n-k+1$ (that is: at least $n-k+1$ of the numbers $x_i$ are zero) by means of the following facts:

If a non-constant polynomial $Q$ has all its roots real, then so does its derivative $Q'$. Then the roots of $Q'$ are the multiple roots of $Q$, plus one root between each pair of consecutive distinct roots of $Q$, necessarily simple (otherwise the sum of the multiplicities of the roots of $Q'$ would exceed its degree). In particular, if $x$ is a multiple root of $Q'$ then it is also necessarily a root of $Q$.