Is the language that consists of machine configurations whose language is a subset of even palindromes semi-decidable? Let $PAL = \{ww^R\ | w\in\{0,1\}^*\}$. 

Then let $A = \{\langle M\rangle \ | \textit{M is a Turing Machine and } L(M)\subseteq PAL\}$

Is A semi-decidable (Turing recognizable or recursively enumerable)?

My try!

I said it is so here is my proof, tell me if I am way off base or tell me if A ought not to semi-decidable in the first place.

$\underline{Proof}$

A is semi-decidable by The Certificate Theorem, which states that:

If $A\subseteq\Sigma^*$ then $A\in SD$(semi-decidable) iff there is some decidable relation, $R\subseteq \Sigma^* \times \Sigma^*$, such that $\forall x\in\Sigma^*$, $x\in{A}$ iff $\exists y$ such that $R(x,y)$ holds.

We know that if $x\in{A}$ that means that $x=\langle M\rangle$, so we can choose $y = \langle M,w\rangle$ where $w\in{PAL}$. What we know is that when $M$ takes $w$ as an input it will accept (since we choose $w$ as an input), if that is the case then $w\in{L(M)}\subseteq{PAL}$. That means that $A\in{SD}$. QED

Any help would be greatly appreciated, have a great day!
What you have shown is that the language of Turing machines accepting at least one even palindrome is semi-decidable. However, your question is interested in the language of Turing machines accepting only even palindromes, which is rather different. The complement of this language is the language of Turing machines accepting at least one word which isn't an even palindrome, and so the complement is semi-decidable according to your argument. This still leaves open the question of whether the original language is semi-decidable (and so decidable).