$A \implies B \lor C$, where $B$ is not guaranteed to be true. Can one say $C$ is always true? $A \implies B \lor C$, where $B$ is not guaranteed to be true. Can one say $C$ is always true?
Can I argue that $C$ needs to hold even if $B$ does not hold to prove $C$?
Edit: I will include the precise problem. 
Let it be true that for every subgraph $H \subseteq G$ there exists a set $X\subseteq V(H)$ of pairwise non-adjancent vertices such that $|X|\geq \frac{|V(H)|}{2}$.
Consider all vertices $v_i \in V(G)$ and two sets $M_1, M_2$. Move all vertices from $V(G)$ to $M_1$ and let $M_2$ be empty. Now run through all $v_i \in M_1$ . For a fixed $v_i$, run through all $v_j, j \neq i$. If $v_i$ and $v_j$ are adjancent, move $v_j$ to $M_2$. After running through all $i$, all vertices in $M_1$ are pairwise non-adjancent. 
Now I want to show that no pair of vertices in $M_2$ is adjancent. For that I use the assumption I made about existence of $X$. Run through all $v_k \in M_2$. For a fixed $v_k$ consider a $v_{m} \in M_1$ such that $v_k$ and $v_m$ are adjancent ($v_m$ exists since $v_k$ has been moved to $M_2$ due to this connection). Now run through all $v_n \in M_2$ and consider the subgraph $H$ with $V(H)=\{ v_k, v_m, v_n \}$. Now using the assumption, $|X|=2$, since $v_k, v_m$ are adjacent. Now it follows that $X=\{v_k, v_n\}$, meaning that all pairs in $M_2$ are non-adjancent or $X= \{v_m, v_n\}$, which does not always apply, since there can be a graph $G$ with an edge connecting $v_m, v_n$. Does this prove that $X=\{v_k, v_n\}$?
 A: If I understand you correctly neither is true. As @DerekElkins points out, if you're using terminology such as "always," you probably want first order quantification. In particular, you're probably looking for something like the following statement:
$$(\forall x. A(x) \implies B(x) \lor C(x)) \land (\exists y. \lnot B(y)) \implies (\forall z. A(z) \implies C(z))$$
Unfortunately this is not true and @DerekElkins's example is again a good one: all integers are either even or odd. Just because not all integers are even doesn't mean all integers are odd.
The specific proposition you're trying to prove, namely that $M_2$ as you've constructed must have pairwise disconnected vertices is also false. In particular, consider the "straight line" four-vertex graph, i.e. where $V = \{A, B, C, D\}$ with edges $AB, BC, CD$. 
If I follow your construction, then one possible result is to start at $A$, which moves $B$ to $M_2$. Then I examine $D$, moving $C$ to $M_2$. Thus $M_2$ consists of $\{B, C\}$.
