As you pointed out we have $$tr(V^T A V) = \sum_{i=1}^k \langle A v_i, v_i \rangle$$

Our restraints are $\langle v_i, v_i \rangle = 1$ and $\langle v_i, v_j \rangle = 0$.

Now suppose we took a greedy approach, and tried to maximize $\langle A v_1, v_1 \rangle$ first, and then maximimize $\langle A v_2, v_2 \rangle$ subject to the constraint that $\langle v_1, v_2 \rangle = 0$ and so on. Then by min-max C-F, we would have $\langle A v_1, v_1\rangle = \lambda_1$, and $\langle A v_i, v_i \rangle \geq \lambda_i$ for each $i$. Hence this approach would yield the fact that $\max tr(V^T A V) \geq \sum_1^k \lambda_i$.

This reduces the problem to showing that $tr(V^T A V) \leq \sum_1^k \lambda_i$ for all choices of $v_i$.

As you pointed out we have $$tr(V^T A V) = \sum_{i=1}^k \langle A v_i, v_i \rangle$$

Our restraints are $\langle v_i, v_i \rangle = 1$ and $\langle v_i, v_j \rangle = 0$.

Now suppose we took a greedy approach, and tried to maximize $\langle A v_1, v_1 \rangle$ first, and then maximimize $\langle A v_2, v_2 \rangle$ subject to the constraint that $\langle v_1, v_2 \rangle = 0$ and so on. Then by min-max C-F, we would have $\langle A v_1, v_1\rangle = \lambda_1$, and $\langle A v_i, v_i \rangle \geq \lambda_i$ for each $i$. Hence this approach would yield the fact that $\max tr(V^T A V) \geq \sum_1^k \lambda_i$.

This reduces the problem to showing that $tr(V^T A V) \leq \sum_1^k \lambda_i$ for all choices of $v_i$. This is equivalent to the fact that the diagonal entries of $A$ are majorized by the eigenvalues of $A$, a statement proved in https://en.wikipedia.org/wiki/Schur%E2%80%93Horn_theorem.

I am not aware of a version of the second fact that relies on C-F, although it might exist.