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I wrote a proof of this (and yes, I proved the opposite direction, but I don't have a question about that portion), and I just want to get confirmation that I am not missing anything -- or advice on how to clean it up if it needs that.

Here's the proof:

Suppose W is not a subset of U.

Then there exists vector w belonging to W such that vector w does not ∈ U.

Since U∪W is a subspace, there exists vector u belonging to U such that (vector u + vector w ) ∈ U∪W.

then since W and U are each subspaces, (vector u + vector w)∈ W, and (vector u + vector w)∈ U.

Therefore U is a subset of W.

Can I just end it there?

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    "Since U∪W is a subspace, there exists vector u belonging to U such that (vector u + vector w ) ∈ U∪W. " It's trivially true for all $u\in U$, so I don't really understand what you are trying to say here. "then since W and U are each subspaces, (vector u + vector w)∈ W, and (vector u + vector w)∈ U." - how does this follow from the last sentence? Also, there is no reason for $u+w$ to belong to $U$ because $w$ does not belong to $U$. – 2017-01-25
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    Since subspaces must have closure under vector addition, U, W and UunionW must all have (u+w) in them. I think I said it poorly, but that's what I am trying to get at – 2017-01-25
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    it's true that if $U\cup W$ is a subspace then $u+w\in U\cup W$ because $u,w\in U\cup W$ and subspace is closed under vector addition. But that does not mean $u+w\in U$, for example, because $w$ is not an element of $U$ so closure under vector addition asserts nothing about $u+w$. In fact, by your first assumption that $w\notin U$, you can cross out the case $u+w\in U$ completely. Can you see why? – 2017-01-25
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    Ah, I think so. because w does not belong to U. So after saying that u+w∈U∪W, I should say that since w does not belong to U, u must belong to W to maintain closure. – 2017-01-25
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    That's kind of correct, I guess better way of saying is this; If $u+w$ belong to $U$, then for some $u'\in U$, $u+w=u'$, so that $w=u'-u$. But since both $-u,u'$ are in $U$, by closure under addition it says that $w=u'-u\in U$. But we assumed $w\notin U$, so by contradiction it follows that $u+w\notin U$. Now since $u+w\in U\cup W$ this then implies that $u+w\in W$. Can you see then that this implies $u\in W$, so that $U\subset W$? – 2017-01-25
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    I was trying to show that if W is not a subset of U, then U must be a subset of W give the initial conditions. I follow what you said, except how can I claim u+w belongs to U, when I already said that w is not in U? Instead, can I say that u +w belonging to the union must belong to W since it isn't in U, therefore U is a subset of W? – 2017-01-25
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    If you assume $w\notin U$ and initial conditions, then $u+w$ does not belong to $U$; this is a fact, and proof goes as I commented above. Then yes, $u+w$ beloning to $U\cup W$ means it must belong to at least one of $U$ or $W$, but we know that it doesn't belong to $U$, so it must belong to $W$. Then this implies that $u\in W$. So you proved: for all $u\in U$, $u\in W$. This is exactly the definition of $U\subseteq W$. – 2017-01-25

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