[proofplan] We prove the two required conclusions separately. First, every edge $\{u,v\}$ satisfies the LP constraint $x_u+x_v\ge 1$, which forces at least one endpoint to have value at least $\frac{1}{2}$; hence the threshold set $C$ covers every edge. Second, each selected vertex $v\in C$ contributes at least $\frac{1}{2}$ to the LP objective, so summing the inequalities $1\le 2x_v$ over $C$ gives the claimed cardinality bound. [/proofplan] [step:Show that every edge has an endpoint in the rounded set] Let $\{u,v\}\in E$ be an arbitrary edge. Since $x$ is feasible for the vertex cover linear programming relaxation, the edge constraint gives \begin{align*} x_u+x_v\ge 1. \end{align*} Suppose, toward a contradiction, that neither endpoint belongs to $C$. Then $x_u<\frac{1}{2}$ and $x_v<\frac{1}{2}$, so \begin{align*} x_u+x_v<\frac{1}{2}+\frac{1}{2}=1, \end{align*} contradicting $x_u+x_v\ge 1$. Therefore at least one of $u$ or $v$ lies in $C$. Because the edge $\{u,v\}\in E$ was arbitrary, every edge of $G$ has at least one endpoint in $C$. Hence $C$ is a vertex cover of $G$. [guided] We need to verify the defining property of a vertex cover: for every edge, at least one endpoint must be selected. Let $\{u,v\}\in E$ be any edge. The feasibility of $x$ gives the LP constraint \begin{align*} x_u+x_v\ge 1. \end{align*} The rounded set $C$ contains exactly those vertices whose LP value is at least $\frac{1}{2}$. If neither endpoint were selected, then both inequalities \begin{align*} x_u<\frac{1}{2} \quad\text{and}\quad x_v<\frac{1}{2} \end{align*} would hold. Adding these two strict inequalities gives \begin{align*} x_u+x_v<1, \end{align*} which contradicts the edge constraint $x_u+x_v\ge 1$. Thus every edge $\{u,v\}$ has $u\in C$ or $v\in C$. Since the edge was arbitrary, $C$ meets every edge, so $C$ is a vertex cover. [/guided] [/step] [step:Bound the number of selected vertices by twice the LP value] For every vertex $v\in C$, the definition of $C$ gives \begin{align*} x_v\ge \frac{1}{2}. \end{align*} Equivalently, \begin{align*} 1\le 2x_v. \end{align*} Summing this inequality over all $v\in C$ gives \begin{align*} |C| =\sum_{v\in C}1 \le \sum_{v\in C}2x_v =2\sum_{v\in C}x_v. \end{align*} Since $x_v\ge 0$ for every $v\in V$, enlarging the summation domain from $C$ to $V$ can only increase the sum: \begin{align*} 2\sum_{v\in C}x_v\le 2\sum_{v\in V}x_v. \end{align*} Combining the two inequalities yields \begin{align*} |C|\le 2\sum_{v\in V}x_v. \end{align*} This proves both claims. [/step]