complexity theory – $forall Anotin RE$ prove that $L_A ={langle Mrangle : |Abigcap L(M)|ge10 }notin RE $

My solution for this question is:
Reduction from $L_A$ to $A$, in the following way $f(x)=langle M_xrangle$
Emphasis: $exists$ 10 different words $w_1 ,dots,w_{10}in A$, otherwise $A$ finite $Rightarrow Ain R Rightarrow Ain RE$ in contradiction to $Anotin RE$
$M_x$ on input $y$ implemented in the following way:

  1. $M_x$ accept 9 of 10 words $w_1 ,dots,w_{10}$ that different from $y$ (because all the words different at least 9 words that different from $y$ exists)
  2. if $y$ equals to $w_i$, $M_x$ accept

The reduction defined for all imputes, can be realized, validity:
$yin ARightarrow M_x$ accepts 10 words from $ARightarrow langle M_x rangle in L_A$
$ynotin ARightarrow M_x$ accepts 9 words from $ARightarrow langle M_x rangle notin L_A$
$yin ALeftrightarrow langle M_x rangle in L_A$ it means $Ale L_A$, given $Anotin RE$ so $L_A notin RE$
Did I make a mistake somewhere?