$U_1 \not\subseteq U_2$, $U_2 \not\subseteq U_1$ $\implies$ $U_1 \cup U_2$ is not a subspace

Question

Let $K$ be a field and $V$ a $K$-vector space and $U_1, U_2 \subseteq V$ subspaces of $V$ that satisfy $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$. Prove that $U_1 \cup U_2$ is not a subspace of $V$.

I only found the contraposition of this argument here and here.

My counterexample goes as follows:

One obtaines $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$ $\implies U_1 \cap U_2 = \emptyset$. For $U_1 \cup U_2$ to be a subspace of $V$ we have to prove the three subspace axioms, the first being that it's not empty.

But if we choose $$ V := \{ p \in \mathbb{R}[t]: deg(p) < 2 \}, \quad U_1 := \{ p \in V: p(0) = 0 \}, \text{ and } U_2 := \{ p \in V: p(0) = 1 \} $$ We obtain $U_1 \cap U_2 = \emptyset$.

Is my counterexample correct?

Answer 1

Hint for the proof: Choose $u_1\in U_1\backslash U_2$ and $u_2\in U_2\backslash U_1$ prove that $u_1+u_2\not\in U_1\cup U_2$.

If by contradiction $u_1+u_2$ is an element of, say $U_1$ what can you say about $u_2$?

Answer 2

One obtains $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$ $\implies U_1 \cap U_2 = \emptyset$.

No. For one thing $0\in U_1\cap U_2$ always.

But if we choose et cetera.

$U_2$ is not a vector subspace, because $0\notin U_2$.

---

In fact your (like any other) counterexample cannot be correct, because the theorem is true.