To elaborate on fedja's comment: Let $(X,\mu)$ be a measure space, let $h$ be a bounded measurable complex-valued function on $X$, and let $T$ be the multiplication operator on $L^2(X,\mu)$ defined by $Tf = hf$. Show that $T$ is normal, and is self-adjoint iff $h$ is real-valued almost everywhere. Now show that $\lambda$ is an eigenvalue of $T$ iff $\mu(\{h= \lambda\}) > 0$. Taking as an example $X = [0,1]$ with Lebesgue measure, you should be able to use this to construct a normal, non-self-adjoint operator with only real eigenvalues, or with no eigenvalues at all.
The correct statement would be if a normal operator has only real spectrum (i.e. $\sigma(T) \subset \mathbb{R}$) then it is self-adjoint. But in infinite dimensions, an operator's spectrum may be more than just its eigenvalues.
Geometrically, it's probably best to think about self-adjoint operators in terms of their eigenspaces. An operator on a finite-dimensional inner product space is self-adjoint if and only if its eigenvalues are real and its eigenspaces are orthogonal and sum (directly) to the whole space.
The real eigenvalues means, roughly, there can't be any kind of rotation happening in any plane. All of the orthogonal spaces must stretch, shrink, and/or reflect.
Here's some examples, and geometric reasoning to support why/why not they are self adjoint:
Rotations in a plane
As stated before, there can't really be rotations while remaining self-adjoint, as these produce complex eigenvalues (of modulus $1$, in fact).
Projections onto a line/plane/subspace by least distance
Yep! These are self-adjoint. In essence, we are decomposing the space into the space we are projecting onto (the range), and its orthogonal complement (the kernel). We are leaving the vectors in the range alone (i.e. multiplying them by $1$), and shrinking the vectors in kernel to nothing (i.e. multiplying them by $0$).
Reflections, by least distance
Also self-adjoint. Rather than shrinking the complement to nothing, instead we are reflecting and multiplying the vectors by $-1$. This still makes them self-adjoint, but it will mean that the map is not positive-(semi)definite.
Projections onto one subspace, along a complementary subspace
This is a more general type of projection, which won't generally be self-adjoint, as the complementary subspace need not be orthogonal to the original subspace.
Hope that helps!
EDIT: Regarding orthogonal eigenspaces, suppose that $T : V \to V$ is self-adjoint, and $v_1, v_2$ are eigenvalues for distinct eigenvalues $\lambda_1, \lambda_2$. We simply need to show $\langle v_1, v_2 \rangle = 0$.
To prove this, consider
\begin{align*}
\lambda_1 \langle v_1, v_2 \rangle &= \langle \lambda_1 v_1, v_2 \rangle \\
&= \langle Tv_1, v_2 \rangle \\
&= \langle v_1, Tv_2 \rangle \\
&= \langle v_1, \lambda_2 v_2 \rangle \\
&= \overline{\lambda_2} \langle v_1, v_2 \rangle \\
&= \lambda_2 \langle v_1, v_2 \rangle,
\end{align*}
where the last line uses the fact that $\lambda_2$ is real. Thus, we have
$$(\lambda_1 - \lambda_2)\langle v_1, v_2 \rangle = 0 \implies \langle v_1, v_2 \rangle = 0$$
since $\lambda_1 - \lambda_2 \neq 0$.
Best Answer
The identity matrix is self-adjoint and all of its eigenvalues are equal (to $1$). The problem is in your false understanding that for a matrix to be diagonalizable, it has to have $n$ distinct eigenvalues ($n$ being the relevant dimension). That is in incorrect, as the identity matrix (and many others) show. You are probably confused with the true statement that if an $n\times n$ matrix has $n$ distinct eigenvalues, then it is diagonalizable.