Prove the contrapositive. If $B$ is not diagonalizable, it is of the form $SJS^{-1}$, where $J$ is a size-2 Jordan block. So if $B$ also has positive eigenvalues, $B^2=SJ^2S^{-1}$ will not be diagonalizable.
However, the rest of your proof isn't totally valid. In particular, the identity matrix satisfies all your conditions, but has a lot more than 4 diagonalizable square roots...
EDIT: If $A$ has distinct eigenvalues, then $\sqrt{A}$ does as well, which means it's guaranteed to be diagonalizable.
Here are some worked examples that provide an answer to Question 2. The seminormal matrices in $\text{Mat}_{2\times 2}(\mathbb{K})$ are the symmetric matrices and matrices of the form
$$T(a,b):=\begin{bmatrix}a&b\\-b&a
\end{bmatrix}\,,$$ where $a$ and $b$ are elements of $\mathbb{K}$. For a symmetric matrix $$S(a,b,d):=\begin{bmatrix}a&b\\b&d\end{bmatrix}\,,$$
it is diagonalizable over $\mathbb{K}$ if and only if $a=d$ and $b=0$, or the quadratic polynomial $$x^2+(a+d)\,x+(ad-b^2)\in\mathbb{K}[x]$$ has two distinct roots in $\mathbb{K}$ (if $\text{char}(K)\neq 2$, the second condition is equivalent to stating that $$\Delta(a,b,d):=\sqrt{\left(\dfrac{a-d}{2}\right)^2+b^2}\in\bar{\mathbb{K}}$$ is a nonzero element of $\mathbb{K}$). It turns out that, if $S(a,b,d)$ is diagonalizable over $\mathbb{K}$, then
- when $\mathbb{K}$ is of characteristic $2$, $S(a,b,d)$ is also orthogonally diagonalizable over $\mathbb{K}$; and
- when $\mathbb{K}$ has characteristic not equal to $2$, $S(a,b,d)$ is orthogonally diagonalizable over $\mathbb{K}$ if and only if $a=d$ and $b=0$, or $\mathbb{K}$ contains both $\Delta(a,b,d)$ and
$$\Xi(a,b,d):=\sqrt{2\,\Delta(a,b,d)\,\left(\Delta(a,b,d)-\frac{a-d}{2}\right)}\in\bar{\mathbb{K}}\,.$$
This provides a counterexample to Question 2. For example, when $\mathbb{K}$ is the field of rational numbers $\mathbb{Q}$, we can take $(a,b,d):=(6,4,0)$, so that $\Delta(6,4,0)=5$ and $\Xi(6,4,0)=2\sqrt{5}\notin\mathbb{Q}$. Therefore, $$S(6,4,0)=\begin{bmatrix}6&4\\4&0\end{bmatrix}$$ is not orthogonally diagonalizable over $\mathbb{Q}$. However, $S(6,4,0)$ is diagonalizable over $\mathbb{Q}$ because $\Delta(6,4,0)=5\in\mathbb{Q}_{\neq 0}$.
The smallest subfield $\mathbb{K}$ of $\mathbb{R}$ such that any matrix $S(a,b,d)$, with $a,b,d\in\mathbb{K}$, which is diagonalizable over $\mathbb{K}$, is always also orthogonally diagonalizable over $\mathbb{K}$ is the field of constructible real numbers. Over this field, $S(6,4,0)$ is no longer a counterexample. The same can be said for any field $\mathbb{K}$ that contains all of its square roots (that is, if $S(a,b,d)$ is diagonalizable over $\mathbb{K}$, then it is also orthogonally diagonalizable).
Now, we analyze $T(a,b)$. If $\text{char}(\mathbb{K})=2$, then $T(a,b)$ is diagonalizable over $\mathbb{K}$ if and only if $b=0$, in which case $T(a,b)$ is also orthogonally diagonalizable. If $\text{char}(\mathbb{K})\neq 2$, then $T(a,b)$ is diagonalizable over $\mathbb{K}$ if and only if $b=0$ or $\sqrt{-1}\in\mathbb{K}$; however, when $b\neq 0$, $T(a,b)$ is never orthogonally diagonalizable over $\mathbb{K}$, even when $\mathbb{K}$ contains $\sqrt{-1}$, because it is not symmetric. Unfortunately, the eigenspaces of both $T(a,b)$ and $\big(T(a,b)\big)^\top$ are identical:
$$\mathbb{K}\,\begin{bmatrix}1\\+\sqrt{-1}\end{bmatrix}\text{ and }\mathbb{K}\,\begin{bmatrix}1\\-\sqrt{-1}\end{bmatrix}\,.$$
The counterexamples for $\text{Mat}_{2\times 2}(\mathbb{K})$ (for Question 2) above can be extended to counterexamples for $\text{Mat}_{n\times n}(\mathbb{K})$ whenever $n>2$. Both $S(a,b,d)$ and $T(a,b)$ so far, even when they are diagonalizable over $\mathbb{K}$ but not orthogonally diagonalizable over $\mathbb{K}$, do not provide a counterexample for Question 3.
Best Answer
Every normal matrix is (unitarily) diagonalizable (a standard result), but not every diagonalizable matrix is normal.
If $S$ is an invertible matrix that is not unitary (i.e., $S^* \ne S^{-1}$) and $D$ is a diagonal matrix that does not commute with $S$, then $A := SDS^{-1}$ is diagonalizable but not normal.