Consider the exponential family $$p_\theta(dy) = h(y)\exp(\langle \theta,T(y)\rangle - \Phi(\theta))\mu(dy)$$ with $\theta \in \mathbb R^d$ and the partition function $\Phi(\theta)<\infty$ for all $\theta \in \mathbb R^d$. Suppose $\nabla \Phi: (\mathbb R^d,\|\cdot\|_2) \to (\mathbb R^d,\|\cdot\|_2)$ is L-Lipschitz.
Fix any $v$ on the $d-1$ dimensional sphere. Then it can be shown $X = \langle v, T(Y)\rangle$ is $2L$ sub-Gaussian.
Now, consider the family $$p_\theta(dy) = \exp(-(y-\mu)^2/2)/\sqrt{2\pi}\,dy$$ This is an exponential family with $\theta = [\mu \mid \mu^2]$ and $T(y) = [y\mid -1/2]$. But then this implies $$X = \langle e_1,T(Y)\rangle = Y$$ is $2L$ sub-Gaussian for every $L>0$ (since $\Phi(\theta) \equiv \log \sqrt{2\pi}$)
In a similar vein, by taking $\theta = [\mu \mid \mu^2 \mid -1/2]$ and $T(y) = [y \mid -1/2 \mid y^2]$, one can deduce the problematic statements that $Y^2$ is 1) sub-Gaussian, and 2) has zero variance.
Where is the error?
In your first subcase, why is $\mu^2$ a parameter while it is not multiplied by any $y$? It seems that in your parameterization $\Phi$ depends implicitly on $y$. To see it explicitly: $\Phi(\mu) = .5\log 2\pi = -\langle e_2,T(y) \rangle \log 2\pi$. The intuition is that $T(y)$ must be a vector of sufficient statistics, which obviously a constant is not.
A correct parameterization could be $\theta=\mu,T(y)=y, h(y)=\exp(-.5\times y^2)$ and $\Phi(\theta)=.5(\log2\pi+\mu^2)$. However, we then have $\nabla\Phi=\mu$, which is $L\geq 1$ Lipschitz, which makes sense for a Gaussian distribution with variance 1.
Edited: corrected for the problems pointed in the discussion.