The error is in stating $I(X;Y) = h(X)$, in fact $I(X;Y)$ isn't defined when $X$ is a continuous random variable and $Y=X$ because the random (vector) variable $(X,Y)$ isn't continuous (doesn't have a joint pdf).
The joint cdf $F(x,y) = P(X \leq x, Y \leq y) = P(X \leq \min(x,y))$ has a "ridge" along $x=y$ which prevents it from being differentiable. Formally the "density" would be $f_X(x) \delta(x-y)$ where $\delta$ is the dirac distribution (generalized function), but it makes no sense to plug this in into the integral formula for mutual information of continuous variables since the log of a distribution (generalized function) and product of distributions (generalized function) with non-disjoint support isn't defined.
What you can do is to consider $X$ with a discrete distribution with $2^n$ equiprobable values in $[0,1]$, then $I(X;Y) = H(X) = n$. Then you get that the supremum over those distributions is $\infty$ and thus $C = \infty$.
You can relate differential entropy to classical entropy by considering a quantization of the continuous random variable, see my answer here Differential Entropy
This interpretation says that the entropy of the quantized variable is approximately (in fine quantization limit) equal to the differential entropy plus the entropy of a uniform variable with $2^n$ values if quantization is done uniformly with intervals of length $\Delta = 2^{-n}$. In case of $(X,Y)$ being continuous this means that the continuous mutual information $I(X;Y) = h(X)-h(X|Y)$ both terms have the additional $\log \Delta$ terms when comparing to the discrete mutual information of the quantized versions of the variables. This motivates the usage of continuous mutual information as a fine quantization limit or quantization independent measure.
Edit:
That $I(X;Y)$ isn't defined when $(X,Y)$ doesn't have a pdf might be a bit of a cop-out. You may also consider the channel $Y=X$ as the limit of channels $Y_\sigma = X + Z_\sigma$ where $Z_\sigma \sim N(0,\sigma)$ and $X$ and $Z_\sigma$ are independent. The limit considered is when $\sigma \to 0$.
Then $I(X;Y_\sigma) = h(Y_\sigma) - h(Y_\sigma|X)$. The pdf of $Y_\sigma$ is the convolution of the pdf of $X$ and $Z_\sigma$, so in case $X$ is uniform on $[0,1]$ it will be a mollified (smoothed) version of the uniform distribution on $[0,1]$ and will tend to the uniform distribution as $\sigma \to 0$. So $h(Y_\sigma) \to 0$ as $\sigma \to 0$. The distribution of $Y_\sigma$ conditioned on $X=x$ will be $N(x,\sigma)$ and thus have differential entropy $\log(\sigma \sqrt{2\pi e})$ which is independent of $x$ so integrating it multiplied with the pdf of $X$ gives $h(Y_\sigma | X) = \log(\sigma \sqrt{2\pi e}) \to -\infty$ and thus $C_\sigma = \sup_X h(Y_\sigma)-h(Y_\sigma|X) \to \infty$ as $\sigma \to 0$.
This indicates that $h(X|X)$ might be interpreted as $-\infty$ which certainly isn't 0. This is also consistent with $h(Z_\sigma) \to -\infty$ as $\sigma \to 0$ which is considering a constant random variable approximated by gaussians. The same happens if you consider uniform distribution $U_n$ on $[0,2^{-n}]$ where $h(U_n) = -n \to -\infty$, etc.
However, beware of overinterpreting this. You may also consider an approximation $D_{m,n}$ of a discrete variable with $2^n$ equiprobable values in $[0,1]$ by placing intervals of width $2^{-m}$ (where $m\geq n$) covering each of the $2^n$ values. Then $h(D_{m,n}) = n-m \to -\infty$ as $m\to \infty$. So making the continuous variable approach a discrete distribution makes its differential entropy tend to $-\infty$.
So, just because differential entropy "is" $-\infty$ doesn't mean the random variable is constant. This also illustrates another aspect of the difference between classical entropy and differential entropy.
