We have to the $4$ step check in order to check if it is a subring.
$0 \in C(X)$ because $x\cdot0 = 0\cdot x$.
$1 \in C(X)$ because $x\cdot 1 = 1\cdot x$.
Now we have to prove closure with two operations. Let $c_1 \in C(X)$ and $c_2 \in C(X)$. Now, we know that,
\begin{align} c_1 &= xc_1x^{-1} \\ c_2 &= xc_2x^{-1} \end{align} So, we must show that $c_1 + c_2 \in C(X)$. Now,
\begin{align*} c_1 + c_2 &= xc_1x^{-1} + xc_2x^{-1} \\ &= x(c_1x^{-1} + c_2x^{-1}) \\ &= x(c_1 + c_2)x^{-1} \end{align*} Therefore, $c_1 + c_2$ is in $C(X)$. We must also show that $c_1c_2$ is in $C(X)$. Now, \begin{align*} c_1c_2 &= xc_1x^{-1}xc_2x^{-1} \\ &= xc_1c_2x^{-1} \end{align*} Hence, $c_1c_2$ is in $C(X)$. Since all four conditions for subring has been passed, the ring $C(X)$ is a subring of $R$.
This was my solution but apparently it is wrong. How do we solve this problem?
For sums: Let $x$ be arbitrary, $c_1, c_2 \in C(X)$, then:
$$(c_1+c_2)x= c_1x+ c_2x = xc_1 + xc_2 = x(c_1+c_2)$$ using standard distributive laws, so by definition $c_1+c_2 \in C(X)$ because it commutes with every $x$.
You cannot assume $c_1^{-1}$ nor $c_2^{-1}$ exists as you do. What's given is it commutes with every $x$ and that's also the condition to be checked.
Similarly
$$(c_1c_2)x = c_1(c_2x)=c_1(xc_2) = (c_1x)c_2 = (xc_1)c_2 = x(c_1c_2)$$
for all $x$ and so $c_1c_2 \in C(X)$. And negation is easy too:
$$(-c_1)x= -(c_1x)=-(xc_1) = (xc_1)\cdot (-1) = x(c_1 \cdot (-1)) = x(-c_1)$$ or do substraction in one go as we did addition.