Let $V$ be a real vector space, and let $W_1,W_2 \subseteq V$ be subspaces of $V$. Let $$ W = \{\mathbf v_1 + \mathbf v_2 \mid \mathbf v_1 \in W_1 \text{ and } \mathbf v_2 \in W_2\}. $$ Prove that $W$ is a subspace of $V$.
Typically I would prove the three axioms that define a subspace, but I cannot figure out how to do that for this problem. Any help appreciated!
More explicitly...
Is $0$ an element of $W$?
Yes, because since $W_1$ and $W_2$ are both subspaces, they each contain $0$ themselves and so by letting $v_1=0\in W_1$ and $v_2=0\in W_2$ we can write $0=v_1+v_2$. Since $0$ can be written in the form $v_1+v_2$ with $v_1\in W_1$ and $v_2\in W_2$ it follows that $0\in W$.
If $\alpha$ is an arbitrary scalar and $w\in W$ does it follow that $\alpha w\in W$?
Yes, because since we are told $w\in W$ we know that there exists some $v_1\in W_1$ and $v_2\in W_2$ such that $w=v_1+v_2$.
Then notice that $\alpha w = \alpha(v_1+v_2)=(\alpha v_1)+(\alpha v_2)$ and further notice that $(\alpha v_1)\in W_1$ and $(\alpha v_2)\in W_2$ because $W_1$ and $W_2$ are both subspaces themselves. In other words, we have written $\alpha w$ as a sum of two vectors, the first of which is from $W_1$ and the second of which is from $W_2$, which is precisely the requirement to be an element of $W$, therefore $\alpha w\in W$.
If $w\in W$ and $x\in W$ does it follow that $w+x\in W$?
Yes, because since we are told that $w\in W$ we know there exists some $v_1\in W_1$ and $v_2\in W_2$ such that $w=v_1+v_2$.
We also know that since we are told that $x\in W$ that there exists some $y_1\in W_1$ and $y_2\in W_2$ such that $x=y_1+y_2$.
Here, it is common for people to have instead called the second element $w'=y_1'+y_2'$ so that we can reuse as much of the notation as possible to make sure we remember where elements live. The notation using a ' here has no mathematical significance beyond the fact that these are potentially different vectors that $w$ and $w'$ might be different or the same, but need not be related in any way.
We have then that $w+x = (v_1+v_2)+(y_1+y_2)=v_1+v_2+y_1+y_2=v_1+y_1+v_2+y_2=(v_1+y_1)+(v_2+y_2)$
Noticing then that $v_1$ and $y_1$ are both elements of the subspace $W_1$ and since $W_1$ is closed under addition, $(v_1+y_1)$ is an element of $W_1$. Similarly $(v_2+y_2)\in W_2$. Therefore we see that $w+x$ can be written as the sum of two vectors, one of which from $W_1$ and the other of which from $W_2$, which is precisely what is needed to be an element of $W$.
Note: these three properties can all be combined into a single property to check which will save time, effort, and paperspace. You can prove that this one property is true if and only if all three of the aforementioned properties are true. The property is:
If $\alpha$ and $\beta$ are scalars and $w$ and $x$ are both elements from our space, it follows that $\alpha x+ \beta y$ is also an element of our space.
Here, suppose $\alpha$ and $\beta$ are scalars and $w$ and $x$ are both elements of our space.
Then $w=v_1+v_2$ for some $v_1\in W_1$ and $v_2\in W_2$ and $x=y_1+y_2$ for some $y_1\in W_1$ and $y_2\in W_2$.
Since we know $(\alpha v_1+\beta y_1)$ is a linear combination of elements in $W_1$ and $W_1$ is a vector space we know that $(\alpha v_1 + \beta y_1)\in W_1$. Similarly $(\alpha v_2+\beta y_2)\in W_2$. Then $\alpha w + \beta x = \underbrace{(\alpha v_1+\beta y_1)}_{\in W_1}+\underbrace{(\alpha v_2+\beta y_2)}_{\in W_2}$ is able to be written as the sum of two elements, one of which an element of $W_1$ and the other an element of $W_2$ and is therefore also an element of $W$
To see that if $w, w' \in W$ then $w+w'\in W$, recall that, if $w_1\in W$, then you can write $$w=v_1+v_2, v_1\in V_1, v_2\in V_2$$ and the same with $w'$: $$w'=v_1'+v_2', v_1'\in V_1, v_2\in V_2'$$ Now, use that $v_1+v_1'\in V_1$, and $v_2+v_2'\in V_2$. Can you continue from here?