**Update:** *This answers the question without the extra condition on the subspace passing through the centre.*

The answer is yes. The following stronger statement is an elementary fact of polyhedral geometry:

Every polytope is **affinely isomorphic** to the intersection of a simplex with an affine subspace.

One of the most standard useful facts of polyhedral geometry is that every polytope $P$ is the *projection* of a simplex. This is very simple to see: if $(v_i)_{i=1,\ldots,m}$ with $v_i\in\mathbb{R}^n$ are the vertices of $P$, then $P$ consists of precisely those points $x\in\mathbb{R}^m$ which can be written in the form
$$x = \sum_{i=1}^m c_i v_i, \qquad \sum_{i=1}^m c_i = 1, \qquad c_i \geq 0.$$
Here, the simplex is the standard simplex in $\mathbb{R}^m$, as per the second and third equation. The first equation defines the linear projection from $\mathbb{R}^m$ down to $\mathbb{R}^n$. So the vertices of the simplex map bijectively to the vertices of $P$, and this assignment extends uniquely to an affine map.

Now let's see what happens to this statement under duality. The dual of a simplex is again a simplex, and the dual of a *surjective* affine map is an *injective* affine map. Hence we prove the claim upon using $P^{\vee\vee} = P$ together with the fact that $P^{\vee}$ is the projection of a simplex.

We can also use similar reasoning as above to write $P$ directly as the intersection of a positive orthant with an affine subspace, which is easily seen to be equivalent to the claim. This time, we should use the $H$-representation of $P$. So there is a matrix $A\in\mathbb{R}^{k\times n}$ and a vector $b\in\mathbb{R}^k$ such that $P$ is the set of solutions of the linear system
$$A x - b \geq 0.$$
If $P$ is full-dimensional, then $k$ is the number of facets. This inequality shows that $P$ is the intersection of the positive orthant $\mathbb{R}_+^k$ with the affine subspace parametrized by $Ax - b$.