If $\succsim$ is transitive but irreflexive, then it is asymmetric, proof If $\succsim$ is transitive but irreflexive, then it is asymmetric.

this is my proof:  

Suppose $\succsim$ is not asymmetric, which means that for any $x,y \in X$ $x\succsim y \rightarrow y \succsim x$. By definition $\succsim$ is transitive, i.e. for any $x, y, z \in X$ we have $x \succsim y$ &amp; $y\succsim z$ $\rightarrow$ $x \succsim z$. So since we are supposing $\succsim$ is symmetric:  $y\succsim x \rightarrow x \succsim$ y. Now since  $\succsim$  is transitive  $y \succsim x$ &amp;  $x\succsim y \rightarrow y \succsim y$ but  $\succsim$  is irreflexive so it's a contradition and  $\succsim$ is asymmetric.

What do you think of my proof, is it wrong or not clear enough?
A relation that is not asymmetric need not be symmetric. But if $\succsim$ is not asymmetric, there must be elements $y$ and $x$ such that both $x\succsim y$ and $y\succsim x$ hold and these elements are all you need to arrive at a contradiction. 

Also, $x\succsim y\implies y\succsim x$ can hold for an asymmetric relation, namely when $x\succsim y$ does not hold. 

I'd like also to point out that "any" is ambiguous here and might be read as "all" instead
of the intended "some."