Is my proof correct? Positiveness of eigenvalues of positive semi-definite matrix This is a proof from Problem Set #0 (problem 3c, the last one) in the Machine Learning course (CS229) from Stanford (Fall of 2016) (link).

Show that if $A \in \mathbb{R}^{n \times n}$ is positive semi-definite ($A \succcurlyeq 0$), then all its eigenvalues $\lambda_i(A) \geq 0$

The proof from the official solutions
Let $x \in \mathbb{R}^n$ be any vector. We know that $A = A^T$ , so that $A = U \Lambda U^T$ for an orthogonal matrix $U \in \mathbb{R}^{n\times n}$ by the spectral theorem. Take the $i$th eigenvector $u^{(i)}$. Then we have $$U^T u^{(i)} = e^{(i)},$$ the $i$th standard basis vector. Using this, we have $$0 \leq {u^{(i)}}^TAu^{(i)} = (U^Tu^{(i)})^T\Lambda U^Tu^{(i)} = {e^{(i)}}^T\Lambda e^{(i)} = \lambda_i(A)$$
Thus, $\lambda_i(A) \geq 0$.
I can follow this proof, and have no problem with it.
My proof
We know $A = A^T$, so $A = U\Lambda U^T$ for an orthogonal matrix $U \in \mathbb{R}^{n\times n}$ by the spectral theorem.
On the other hand, in a previous problem (2c), it is proven that, for any $B \in \mathbb{R}^{m \times n}$, it holds that $$BAB^T \succcurlyeq 0.$$
This means that
$$U^TAU \succcurlyeq 0$$
$$\Lambda \succcurlyeq 0$$
$${e^{(k)}}^T \Lambda e^{(k)} \geq 0$$
$$\sum_{i=0}^{n} \lambda_i {e^{(k)}}_i^2 = \lambda_k \geq 0$$
I think it's expressing the same thing, but I can't help but feel awkward about it, like I'm assuming something that is not necessarily true. I thought that the given proof would use the
$BAB^T \succcurlyeq 0$ fact from earlier, so it feels weird that it didn't (explicitly, at least).
Do you think my proof is well formed?
 A: Your proof seems to me to be to be essentially the same as the one you were given.
On the other hand, why not just say this:  if $A$ has an eigenvalue 
$\lambda < 0, \tag 1$
then
$\exists 0 \ne \vec v \in \Bbb R^n \tag 2$
with
$A \vec v = \lambda \vec v; \tag 3$
then
$\langle \vec v, A \vec v \rangle = \langle \vec v, \lambda \vec v \rangle = \lambda \langle \vec v,  \vec v \rangle < 0, \tag 4$
since
$\langle \vec v, \vec v \rangle > 0; \tag 5$
hence $A$ cannot be positive semi-definite; thus, (1) must be false.
Essentially the same argument in any event.
A: Following the definition (as copied from the OP's link)
A matrix $A\in\mathbb R^{n\times n}$ is positive semi-definite (PSD),
denoted $A\succcurlyeq 0$, if $A=A^T$ and $x^T\!Ax\geqslant 0$
for all $x\in\mathbb R^n$.
there's the conceptional argument:
Assume $\,Ax=\lambda x\,$ with $x$ being a unit vector w.l.o.g., then
$\,0\leqslant x^T\!Ax = \lambda\, x^T\!x = \lambda\,$.
Admittedly, this is not an answer to the formulated question, strictly speaking writing.
