I have been told about this result as a consequence of the Courant-Fischer theorem (with no proof given):
If $A$ is a $n \times n$ symmetric matrix and $U$ is a $n \times r$ matrix $(r \le n)$ such that $U^TU = I$, then $\lambda_{k+n-r}(A) \le \lambda_k(U^TAU) \le \lambda_k(A)$, for $k = 1,\dots,r$.
There, $\lambda_k(M)$ denotes the $k$-th largest eigenvalue of the symmetric matrix $M$. It is told as a consequence of the Courant-Fischer theorem, but I'm not sure how can I prove it directly from it. Looking for information, I knew about the interlacing Cauchy theorem, which seems too more similar to this problem. But we haven't studied this theorem.
So, my question is if there is a way to prove this result directly using Courant-Fischer theorem. If that's only possible going through interlacing theorem, I have still a doubt, because to use it $U^TAU$ must be similar to a submatrix of $A$, but is that true?