To give more context for possible answers: at the very least, a counterexample exists if we drop the assumption that $A,B,C$ be all irreducible (arisen in conversations with Harald). In $\mathbb{A}^{5}$, take
\begin{align*}
A & =\{x_{4}=0,x_{3}^{2}-x_{5}-1=0\}, \\
B & =\{x_{5}=0\}, \\
C & =\{x_{1}^{2}-x_{2}^{2}x_{3}^{2}=0,x_{1}-x_{2}x_{3}+x_{4}=0\}, \\
V_{1} & =\{x_{1}+x_{2}x_{3}=0,-2x_{2}x_{3}+x_{4}=0,x_{5}=0\}, \\
V & =\{x_{1}=x_{2}=x_{4}=x_{5}=0,x_{3}=1\}.
\end{align*}
Then the two candidates for $W_{1}$ are $W_{1}=\{x_{4}=x_{5}=0,x_{3}=1\}$ and $W_{1}=\{x_{4}=x_{5}=0,x_{3}=-1\}$: the second one does not contain $V$, while the first one gives
\begin{equation*}
W_{1}\cap C=\{x_{1}-x_{2}=0,x_{3}=1,x_{4}=x_{5}=0\},
\end{equation*}
which is irreducible already. In this example, $A,B$ are irreducible but $C$ is not.