Here is a proof that uses directly the [separating hyperplane theorem][1], which states the following.

Given a matrix $A$ and column matrix $b$,
\begin{align}
Ax=b, &\text{ for some column matrix }x\ge0 \\
&\iff \\
y^TA\ge0 &\text{ for some column matrix }y \implies y^Tb\ge0
\end{align}

-----

We put the required proposition into the above form with the exception that the matrix is transposed. The premise of the proposition is
\begin{align}
&\begin{bmatrix}
A & b \\
0 & 1
\end{bmatrix}\begin{bmatrix}
-xx_1 \\
x_1
\end{bmatrix}\ge 0  \\
&\iff \\
&\begin{cases} Ax\le b \implies a'^Tx\le b' \\
x_1\ge0 
\end{cases}
\\
&\implies \\
&\begin{bmatrix}a'^T &b'\end{bmatrix} \begin{bmatrix}-xx_1 \\x_1 \end{bmatrix}\ge0
\end{align}

The conclusion of the proposition is 
\begin{align}
&\begin{bmatrix}\lambda^T & \lambda_1\end{bmatrix}\begin{bmatrix}
A & b \\
0 & 1
\end{bmatrix}=\begin{bmatrix}a'^T & b'\end{bmatrix}, \text{ for some} \begin{bmatrix}\lambda^T & \lambda_1\end{bmatrix}\ge0 \\
&\iff \\
&\exists \lambda\ge0 \ni
\begin{cases}
\lambda^TA=a'^T \\ 
\lambda^Tb\le b'
\end{cases}
\end{align}

The premise and the conclusion are seen to be equivalent in view of the separating hyperplane theorem as stated above. We have now proved the desired proposition.

  [1]: https://en.wikipedia.org/wiki/Hyperplane_separation_theorem