Is the set $X = \{(x, y, z) ∈ \Bbb R^3: y = 1\}$ a subspace of $ \Bbb R^3$? I am new to linear algebra and math in general. I understand that you need the zero vector  to be in set $X$, you need it closed under vector addition and vector multiplication. Is this closure for the set $X$ itself or in the general vector space.
For instance if I take two arbitrary vectors $(a_1, 1, a_3) \in X$ and $(a_4, 1, a_6)\in X$, then their sum is not in $S$ itself (since second entry would be 2)  but is in $\Bbb R^3$. Does this satisfy closure property of addition or is a counter example?
How is the zero vector property here coming in play?
 A: Your argument correctly shows $X$ is not a subspace.
You state three conditions for $X$ to be a subspace.  If any one of the three fails, you can stop, and conclude it is not a subspace.  (In fact, in this example all three conditions fail.)
A: As is pointed out on the comments, you need to check for the zero vector to guarantee nonemptiness. If you already know your set is not empty, then you can skip this verification and use the fact that $v-v = 0$ together with the other two closure properties to get that zero is a member of your subspace.
Note that you can check for nonemptiness by any other means, but usually just checking for the zero vector is the easiest/quickest way. Furthermore, also note there is only one empty set, so the only way you could find a "subspace" that satisfies the two closure properties but has no zero vector would be by finding some properties which are preserved under addition and scalar multiplication but which are never satisfied by any vector.
An example of this cleverly disguised empty set can be found in in this answer
