Is the complement of $ww^R$ context-free? Identify the language given by $L = \{ x \in (0,1)^* : x \neq ww^R, w \in (0,1)^*\}$. Note: $w^R$ is the reverse of the string $w$.

Closure property can/should be applied only in the cases when the actual language is not specified. Otherwise, you may arrive at wrong conclusions. The language in question is fully specified, so I doubt whether one can directly apply closure property on $ww^R$, which you say, is its complement or vice versa. The given answer is CFL. But I don't seem to have arrived at it satisfactorily.
If you want to show that $L$ is context-free, you can proceed as follows:

Build a push-down automaton $P$ that accepts $L$. $P$ will read the input and put all the read characters on its stack. At some point it guesses (non-deterministically!) that half of the input has been read. Then it reads the remaining input and checks it a against its stack. If $P$ finds a mismatching character it accepts, if it does find not a mismatch, or the guess for the middle of the word was wrong, it rejects.