I will present my attempt at the proof here and would like feedback on it. I also wonder how this relates to the statement that the union of two subgroups is a subgroup if and only if one is contained in the other.
The first part of the proof shows the case when the union is a subspace. The second part includes a case where this does not hold and thus contradicts the statement. Our argument will centre on the conditions for a subspace, with particular emphasis on closure under addition, and I will briefly present the third condition for the first case. We can affirm that the zero $0$ is present within both subspaces and their union, thus satisfying the first condition.
Given two subspaces, $H$ and $K$. Then we will assume that $H \cup K$ is a subspace.
$1)$
Consider two elements $u$ and $w$ such that $u, w \in H \cup K$. This means that $$ [ u \in H \quad \text{or} \quad u \in K \quad \land \quad w \in H \quad \text{or} \quad w \in K ] $$
If $$u \in H \quad \land \quad w \in H $$ $$\implies (u+w) \in H $$ $$\implies (u+w) \in H \cup K$$ Notice that $(u+w) \in H $ holds because H is closed under addition since it is a subspace itself.
Closure under multiplication by scaling condition. For some scalars $c$ and $d$: $$u \in H \quad \land \quad w \in H $$ $$\implies cu \in H \quad \land \quad dw \in H $$ $$\implies (cu+dw) \in H $$ $$\implies (cu+dw) \in H \cup K$$
$2)$
However if $$u \in H \quad \land w \in K \quad then \quad (u+w) \notin H \lor K$$ $$\implies (u+w) \notin H \cup K$$
This should be enough to conclude that the proposition is generally not true even though it might happen to be as we saw in the first example. Regarding the initial statement (that the union of two subgroups is a subgroup if and only if one is contained in the other), I wonder if the first case of my proof where the proposition holds could confirm this. I, however, am reluctant to conclude that it can, given that I considered elements within subspaces rather than subspaces themselves. But I guess one could argue that if, for instance, $w = K$, given that $w \in H$ and $(u+w) \in H$, it would imply that $H$ includes $K$. The first case would then verify the claim. How can the proof be reformulated such that we also end up with this conclusion? Would setting $w = K$ be enough? Thereby making $K$ a subset of $H$...
As stated in the comments, a single counterexample can save a lot of time in disproving a conjecture :)
@zkutch had a great example which I'll expand upon.
We'll take two lines in the $x-y$ plane (i.e., $\mathbb{R^2}$) -- specifically, and to make this super easy to see, let's pick two special subsets of $\mathbb{R^2}$:
$L_x := \{(x,0)\; \forall x \in \mathbb{R}\}, L_y := \{(0,y)\; \forall y \in \mathbb{R}\}$ (i.e., just the $x$ and $y$ axes).
We can show that each is a subspace of $\mathbb{R^2}$:
$\forall u,v \in L_x;\; a,b \in \mathbb{R}:au+bv = (au_1,0)+(bv_1,0) = (au_1+bv_1,0) \in L_x\;\text{ and } \mathbf{0} = 0u+0b \in L_1\; \square$
Same argument for $L_y$.
Now, take let's take $L_{xy} = L_x \cup L_y$ and see if it is a subspace.
(Trivial): We know $\mathbf{0} \in L_{xy}$ since it is in both.
Now, let's take $p_1 = (0,1), p_2= (1,0) \in L_{xy}$ does this hold?
$$au+bv \in L_{xy}\; \forall u,v \in L_{xy};\; \forall a,b \in \mathbb{R}$$
No, it doesn't -- we can select $a=b=1$ and then:
$$p_1 + p_2 = (0,1) + (1,0) = (1,1) \notin L_{xy}$$