So for a linearly independent set $\mathcal V$ of $k$ vectors $\vec v_1,\dots,\vec v_k$ we define a set $\mathcal W$ of $k$ linearly independent vectors $\vec w_1,\dots,\vec w_k$ putting
$$
\vec w_i:=\begin{cases}\vec v_i-\vec v_j,\,\,\,\text{if}\,\,\,i\neq j\\
\vec v_i,\,\,\,\text{otherwise}\end{cases}
$$
and so we prove that
$$
\mathcal S_O(\vec v_1,\dots,\vec v_k)=\mathcal P_O(\vec v_1,\cdots,\vec v_k)\cap\mathcal P_O(\vec w_1,\dots,\vec w_k)
$$
So assuming for semplicity $i=1$ we observe that
$$
\alpha^1\,\vec v_1+\alpha^2\,\vec v_2+\dots+\alpha^k\,\vec v_k=\big(\alpha^1+(\alpha^2+\dots+\alpha^k)-(\alpha^2+\dots+\alpha^k)\big)\,\vec v_1+\alpha^2\,\vec v_2+\dots+\alpha^k\,\vec v_k=\\
(\alpha^1+\dots+\alpha^k)\,\vec v_1+\alpha^2\,(\vec v_2-\vec v_1)+\dots+\alpha^k\,(\vec v_k-\vec v_1)
$$
for any $\alpha^1,\alpha^2,\dots,\alpha^k\in\Bbb R^n$ and thus we conclude that
$$
\mathcal S_O(\vec v_1,\dots,\vec v_k)\subseteq\mathcal P_O(\vec v_1,\cdots,\vec v_k)\cap\mathcal P_O(\vec w_1,\dots,\vec w_k)
$$
since if
$$
\alpha^1+\alpha^2+\dots+\alpha^k\le 1\,\,\,\text{and}\,\,\,\alpha^1,\alpha^2,\dots,\alpha^k\in[0,1]
$$
then it must be
$$
(\alpha^1+\alpha^2+\dots+\alpha^k),\alpha^2,\dots,\alpha^k\in[0,1]
$$
so that any element of $\mathcal S_O(\vec v_1,\dots,\vec v_k)$ is an element of $\mathcal P_O(\vec w_1,\dots,\vec w_k)$ and thus statement follows directely by the inclusion
$$
S_O(\vec v_1,\dots,\vec v_k)\subseteq\mathcal P_O(\vec v_1,\cdots,\vec v_k)
$$
proved in the question. Moreover we observe that
$$
\alpha^1\,\vec v_1+\alpha^2,\vec v_2+\dots+\alpha^k\,\vec v_k=\beta^1\,\vec w_1+\beta^2\,\vec w_2+\dots+\beta^k\,\vec w_k\Rightarrow\\
\alpha^1\,\vec v_1+\alpha^2,\vec v_2+\dots+\alpha^k\,\vec v_k=\beta^1\,\vec v_1+\beta^2\,(\vec v_2-\vec v_1)+\dots+\beta^k\,(\vec v_k-\vec v_1)\Rightarrow\\
\alpha^1\,\vec v_1+\alpha^2,\vec v_2+\dots+\alpha^k\,\vec v_k=\big(\beta^1-(\beta^2+\dots+\beta^k)\big)\,\vec v_1+\beta^2\,\vec v_2+\dots\,\beta^k\,\vec v_k\Rightarrow\begin{cases}\alpha^1=\beta^1-(\beta^2+\dots+\beta^k)\\
\alpha^2=\beta^2\\
\vdots\\
\alpha^k=\beta^k\end{cases}\Rightarrow\alpha^1+\alpha^2+\dots+\alpha^k=\beta^1\le 1
$$
for any $\alpha^1,\alpha^2,\dots,\alpha^k,\beta^1,\beta^2,\dots,\beta^k\in\Bbb R$ and thus we conclude that
$$
\mathcal P_O(\vec v_1,\cdots,\vec v_k)\cap\mathcal P_O(\vec w_1,\dots,\vec w_k)\subseteq \mathcal S_O(\vec v_1,\dots,\vec v_k)
$$
because if
$$
\alpha^1+\alpha^2+\dots+\alpha^k=\beta^1\,\,\,\text{and}\,\,\,\alpha^1,\alpha^2,\dots,\alpha^k,\beta_1,\beta^2,\dots,\beta^k\in[0,1]
$$
then surely
$$
\alpha^1+\alpha^2+\dots+\alpha^k\le 1\,\,\,\text{and}\,\,\,\alpha^1,\alpha^2,\dots,\alpha^k\in[0,1]
$$
so that the statement follows immediately. So we conclude that any simplex is the intersection of most parallelopides which are compact -indeed any parallelopides is homeomorphic to the unitary cube $[0,1]^k$ via an affine homeomorphism- and convex -see here for details. So finally we conclude that any simplex is the intersection of two compact and convex sets so that the statement follows immediately and moreover in such a way it has been even proved that any simplex is path connected and so connected since convexity implies path connectdness and this last implies connectdness.
