1):

consider a = 1, b = 1, x = y = 1.

isn't it an axiom that (a + b)(x,y) = a(x,y) + b(x,y)? what does the above example lead you to conclude?

2): if W_{1}⊆ W_{2}, there is nothing to prove, so we lose no generality by assuming this is NOT the case. thus there is some w_{1}in W_{1}that is NOT in W_{2}.

so let w_{2}be any element of W_{2}. consider w = w_{1}+w_{2}. since w_{1}is in W_{1}, and w_{2}lies in W_{2}, each term lies in W_{1}UW_{2}.

since W_{1}UW_{2}, is a subspace, it is closed under addition hence w = w_{1}+w_{2}is in W_{1}UW_{2}.

we thus have:

w_{1}= w - w_{2}.

since this is NOT in W_{2}, w cannot be in W_{2}either, or else w - w_{2}= w + (-1)w_{2}would be in W_{2}, since W_{2}is a subspace (and closed under vector addition and scalar multiplication).

but since w is in W_{1}UW_{2}, if it is not in W_{2}, it must lie in W_{1}.

so w_{2}= w - w_{1}lies in W_{1}(since W_{1}is a subspace), and since w_{2}was chosen arbitrarily from W_{2}, we have shown if W_{1}is not a subset of W_{2}, then W_{2}is a subset of W_{1}.