Above we showed that the subset $\overset{\circ}{\mathcal E}_k$ of $\mathcal E_k$ whose elements have value in $[0,1)$ with respect the scalar function $g$ is open in $H^k_k$ since it is union of the sets
$$
g^{-1}\big[\,(0,1)\,\big]\cap H^k_k\,\,\,\text{and}\,\,\,C\Big(O,\frac 1k\Big)\cap H^k_k
$$
that are open in $H^k_k$. Now provided that
$$
\xi^1+\dots+\xi^k=1
$$
for any $\xi\in\mathcal E_k$ then surely it must be
$$
\xi^j\gneq0
$$
for any $j=1,\dots,k$ and in particular for now we assume that this happens for the last coordiante. So in this case we define the map $\varphi$ from $\Bbb R^k$ into $\Bbb R^k$ through the equation
$$
\varphi(x):=\big(x^1,\dots,x^{k-1},1-(x^1+\dots+x^{k-1}+x^k)\big)
$$
for any $x\in\Bbb R^k$ and thus we are going to prove that this map can be restricted to a coordinate patch about $\xi\in\mathcal E_k$ whose last coordinate is not zero, i.e. we let to prove that the restriction $\phi$ of $\varphi$ at the set $\overset{\circ}{\mathcal E}_k$ is the researched coordinate patch. First of all we observe that $\varphi$ is a diffeomorphism of $\Bbb R^k$ onto $\Bbb R^k$ being an affine map, i.e. it is compostion of a translation with a linear map between finite dimensional topological vector spaces and both these maps are diffeomorphism. So if we prove that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into an open set of $\mathcal E_k$ containing $\xi$ then we will proved that $\varphi$ is a coordinate patch about $\xi$ so that we let to do this. Now if $x$ is an elemen of $\overset{\circ}{\mathcal F}$ then it must be
$$
x^i\in[0,1)
$$
for each $i=1,\dots,k$ since otherwise it would be
$$
x^1+\dots+x^{i-1}+1+x^{i+1}+\dots+x^k\le x^1+\dots+x^{i-1}+x^i+x^{i+1}+\dots+x^k<1\Rightarrow\\ x^1+\dots+x^{i-1}+x^{i+1}+\dots+x^k<0
$$
that is impossible if $x$ lies in $\overset{\circ}{\mathcal E_k}$ and thus in $H^k_k$ and so we can conclude that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow x\in[0,1)^k\wedge x^1+\dots+x^k\lneq1\Rightarrow\\\varphi(x)\in H^k_k\wedge\varphi^1(x)+\dots+\varphi^k(x)\le1\Rightarrow\varphi(x)\in\mathcal E_k
$$
and this proves that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into $\mathcal E_k$. So if the last coordinate of $\xi$ is not zero and the others are not negative then surely
$$
g(\xi^1,\dots,\xi^{k-1},0)\ge0\,\,\,\text{and}\,\,\,g(\xi^1,\dots,\xi^{k-1},0)=\xi^1+\dots+\xi^{k-1}<\xi^1+\dots+\xi^{k-1}+\xi^k=1
$$
so that $(\xi^1,\dots,\xi^{k-1},0)$ is an element of $\overset{\circ}{\mathcal E}_k$ and in particular it is such that
$$
\varphi(\xi^1,\dots,\xi^{k-1},0)=\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1})\big)=\big(\xi^1,\dots,\xi^{k-1},\xi^k\big)=\xi
$$
and thus it is proved that $\xi$ is an element of $\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]$. Now we previously define the set
$$
\tilde{\mathcal E}_k:=\{x\in\mathcal E_k:x^k\neq 0\}
$$
and thus we let prove that $\varphi$ carries $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we first observe that
$$
x\in\overset{\circ}{\mathcal E}_k\Rightarrow 0\le g(x)\lneq1\Rightarrow0\lneq1-g(x)\le1\Rightarrow\varphi^k(x)>0\Rightarrow\varphi(x)\in\tilde{\mathcal E}_k
$$
and so we conclude that
$$
\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]\subseteq\tilde{\mathcal E}_k
$$
so that we let to prove the opposite inclusion. Now for any element $\xi$ of $\tilde{\mathcal E}_k$ it must be
$$
g(\xi)=1\,\,\,\text{or either}\,\,\,g(\xi)<1
$$
but effectively we just proved above that if $g(\xi)=1$ and $\xi^k>0$ then surely
$$
\xi=\varphi(x)
$$
for any $x\in\overset{\circ}{\mathcal E}$ so that to follow we are supposing that $g(\xi)$ is strictly less than $1$. So we observe that
$$
\begin{cases}\xi\in\tilde{\mathcal E}_k\\
g(\xi)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\\xi^k\gneq0\\
\xi^1+\dots+\xi^{k-1}+\xi^k\lneq1\end{cases}\Rightarrow\\
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq1-\xi^k\lneq1\\
0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow
\begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\
1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\gneq0\\
0\lneq\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\lneq1\end{cases}
$$
and thus we conclude that $\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)$ is an element of $\overset{\circ}{\mathcal E}_k$ and so observing that
$$
\varphi\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\biggl)=\\
\biggl(\xi^1,\dots,\xi^{k-1},1-\Big(\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\Big)\biggl)=\\
\Biggl(\xi^1,\dots,\xi^{k-1},1-\Big((\xi^1+\dots+\xi^{k-1})+1-(\xi^1+\dots+\xi^{k-1})-\xi^k\Big)\Biggl)=\\
\big(\xi^1,\dots,\xi^{k-1},1-(1-\xi^k)\big)=(\xi^1,\dots,\xi^{k-1},\xi^k)=\xi$$
we conclude that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we let to prove that $\tilde{\mathcal E}_k$ is open in $\mathcal E_k$ and we are doing this proving that the set $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\mathcal E_k$, that is it is the intersection of a closed set of $\Bbb R^k$ with $\mathcal E_k$. Now we remember that any symplex is closed (see here for details) so that the $(k-1)$ standard simplex $\mathcal E_{k-1}$ is closed: so observing that
$$
\mathcal E_k\setminus\tilde{\mathcal E}_k=\{x\in\mathcal E_k:x^k=0\}=\\
\{x\in\Bbb R^k:x^1+\dots x^{k-1}\le1\wedge x^i\ge0,\,\forall\,i=1,\dots,(k-1)\wedge x^k=0\}=\\
\{x\in\Bbb R^{k-1}:x^1+\dots+x^{k-1}\le1\wedge x^i\,\forall\,i=1,\dots,(k-1)\ge0\}\times\{0\}=\mathcal E_{k-1}\times\{0\}
$$
we conclude that $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\Bbb R^k$ (ideed it is product of closed set) and so $\mathcal E_k$ too. So we proved the existence of a coordinate patch when the last coordinate of $\xi$ is not zero. Otherwise if the last coordinate of $\xi$ is zero then surely (remember what observed above) there must exist $j=1,\dots,k-1$ such that
$$
\xi^j\neq0
$$
so that let be $\psi$ the diffeormosphism of $\Bbb R^k$ onto $\Bbb R^k$ that interchanges the $j$-th coordinate with the last, i.e.
$$
[\psi(x)](i):=\begin{cases}x^k,\,\,\,\text{if}\,\,\,i=j\\
x^j,\,\,\,\text{if}\,\,\,i=k\\
x^i\,\,\,\text{otherwise}\end{cases}
$$
for any $x\in\Bbb R^k$. Now $\psi$ is an involution that maps $\mathcal E_k$ onto $\mathcal E_k$ and in particular if the last coordinate of $\xi$ is zero then this does not happen for $\psi^{-1}(\xi)$ so that if $\varphi$ is a coordinate patch about $\psi^{-1}(\xi)$ then the composition $\psi\circ\varphi$ of $\psi$ with $\varphi$ is just a coordinate patch about $\xi$ and so the statement finally holds.
So first of all we remember that the product of convex sets is convex too (see here for details) and moreover we remember that any interval of the real line $\Bbb R$ is convex: so any rectangle is convex being product of convex sets and thus the unit cube $[0,1]^k$ is convex. Now any linear map and any translation between t.s.v. preserve the convexity since if $f:V\rightarrow W$ is a such map then
$$
f\big(x+t(y-x)\big)=f(x)+t\big(f(y)-f(x)\big)
$$
for any $x,y\in V$. So we conclude that any $k$-parallelopiped is convex because it is homeomorphic to the unit cube $[0,1]^k$ via the composition of a translation and a linear map.
Anyway it is possible (if it interest) to prove the statement via other argumentations I show to follow.
So if $x,y\in\mathcal P_O(\vec v_1,\dots\vec v_k)$ then there must exist $\xi^i,\eta^i\in[0,1]$ for $i=1,\dots,k$ such that
$$
x=O+\xi^i\vec v_i\,\,\,\text{and}\,\,\,y=O+\eta^i\vec v_i
$$
and thus we have to prove that
$$
x+t\cdot(x-y)=\big(O+\xi^i\vec v_i\big)+t\cdot(\eta^i-\xi^i)\vec v_i\in\mathcal P_O(\vec v_1,\dots,\vec v_k)
$$
for any $t\in[0,1]$. So observing that
$$
\big(O+\xi^i\vec v_i\big)+t\cdot(\eta^i-\xi^i)\vec v_i=O+\big(\xi^i+t\cdot(\eta^i-\xi^i)\big)\vec v_i=O+\big((1-t)\cdot\xi^i+t\cdot\eta^i\big)\vec v_i
$$
for any $t\in[0,1]$ we observe that
$$
\begin{cases}0\le\xi^i,\eta^i\le1\\0\le t\le1\end{cases}\Rightarrow\begin{cases}0\le\xi^i,\eta^i\le1\\0\le t\le 1\\0\le1-t\le1\end{cases}\Rightarrow\\
\begin{cases}(1-t)\cdot\xi^i\ge0\\t\cdot\eta^i\ge0\\t\cdot\eta^i\le t\le1\\(1-t)\cdot\xi^i\le(1-t)\le1\end{cases}\Rightarrow0\le(1-t)\cdot\xi^i+t\cdot\eta^i\le(1-t)+t=1
$$
for any $i=1,\dots, k$ and this proves the statement.
Best Answer
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.