Can I prove this statement like this? The question asks $\forall x \in \Bbb R$, if $x > 3$ then $x^2 > 16$. The solutions tell me to find all intervals/cases for $x$ and prove that for one of the intervals, the hypothesis $x > 3$ is true and the conclusion $x^2 > 16$ is false, and if that case pops up the statement is false, and otherwise true.
But can I just make my proof "let $x = 4$. Then this statement is false. QED."??
I'm just worried about losing marks on these types of questions, I'm told to prove something so that the person who is reading it can understand but I don't know when something is obvious or I need to explain it or not.
 A: Yes, you can show that the statement to be proved is false just showing a counter-example (a counter-example is a case where the statement to be proved doesn't holds).
In your case the statement $4>3\implies 16>16$ is false, so it is not true that the statement $x>3\implies x^2>16$ is true for all $x\in\Bbb R$.
A: Of course it is enough to proof that $\forall x \in \mathbb{R} : x>3 \rightarrow x^2 > 16$ is false showing that for x=4 the statement is false, but the question regards to show if for some interval it is true, so the answer to your question is NO. The interval of x for every x greater than 4 the statement is correct, hence if you answer that the statement is false you don't answer what the question asks. 
However, I think that the question is not very well stated, it seems to have the purpose to make you lost time thinking what is it intended to say.
A: If you show, that  
$\exists x \in \mathbb{R}$, if $x>3$ then $x^2 \not >16$,  
then you already proved the statement is not true, because the statement doesn't hold for $\forall x\in \mathbb{R}$. 
In other words: If we can find one counterexample $(\exists x\in \mathbb{R}: x>3 \implies x^2 \not > 16)$ for which the given statement $(\forall x\in \mathbb{R}: x>3 \implies x^2 > 16)$ is not true, it is not true for all $x$ and from this it immediately follows that the statement itself is not true.
