Let $X$ be a random variable with a probability density function $f$ whose support is a set $\mathcal{X}$. The entropy $h(X)$ or $h(f)$ is defined as
$$ H(X)=\mathrm{E}[-\log (f(X))]=-\int_{\mathcal{X}} f(x) \log f(x) d x $$
Consider $dx$, the counting or Lebesgue measure.
From T. M. Cover, J. A. Thomas - Elements of Information Theory (2006) :
We have for the uniform distribution the following:
Theorem 2.6.4 $H(X) \leq \log |\mathcal{X}|$, where $|\mathcal{X}|$ denotes the number of elements in the range of $X$, with equality if and only $X$ has a uniform distribution over $\mathcal{X}$.
Proof: Let $u(x)=\frac{1}{|\mathcal{X}|}$ be the uniform probability mass function over $\mathcal{X}$, and let $p(x)$ be the probability mass function for $X$. Then $$ D(p \| u)=\sum p(x) \log \frac{p(x)}{u(x)}=\log |\mathcal{X}|-H(X) . $$ Hence by the nonnegativity of relative entropy, $$ 0 \leq D(p \| u)=\log |\mathcal{X}|-H(X) $$
We have for the normal distribution the following:
Theorem 8.6.5 Let the random vector $\mathbf{X} \in \mathbf{R}^n$ have zero mean and covariance $K=E \mathbf{X X}^t$ (i.e., $K_{i j}=E X_i X_j, 1 \leq i, j \leq n$ ). Then $h(\mathbf{X}) \leq$ $\frac{1}{2} \log (2 \pi e)^n|K|$, with equality iff $\mathbf{X} \sim \mathcal{N}(0, K)$.
Does Theroem 2.6.4 holds also for continuous uniform distributions ? (bounded support)
The Theorem 8.6.5 by letting $\mathbf{R}^n$ be the support, increase the set of possible distributions still intuitively I would think the continuous uniform distribution to be the extension of the discrete case ?
Yes, Theorem 2.6.4 holds more generally. It's known as Gibbs' inequality, and is true e.g. for the continuous uniform distribution on any set of finite measure in $\mathbb R^n$.
Theorem 8.6.5 is different in that it fixes the covariance $K$. If $K$ was allowed to vary, there would be no maximum entropy distribution; you can check that e.g. for $X \sim N(0, \sigma^2I)$, the entropy $h(X) \to \infty$ as $\sigma^2 \to \infty$.
Asking what maximises entropy for a fixed covariance $K$ is a fundamentally different question than the unrestricted maximum, which is why you get a qualitatively different result. Even if the support was bounded, fixing $K$ would give you a Gaussian-like distribution, rather than a uniform one.