Geometric Interpretation. Consider a positive definite matrix $A$. It defines the ellipsoid
$${\cal E}_A = \{u: u^T A u \leq 1\}.$$
Note that the correspondence between $A$ and ${\cal E}_A$ is one-to-one. Moreover, $A\succeq B$ if and only if ${\cal E}_B$ contains ${\cal E}_A$. Using this fact, we can give the following characterization.
$C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$.
Similarly,
$C \preceq A$ and $C \preceq B$ if and only if ${\cal E}_C \supset {\cal E}_A \cup {\cal E}_B$.
Now a matrix $C$ is a minimal matrix s.t. $C \succeq A$ and $C \succeq B$ if and only if ${\cal E}_C \subset {\cal E}_A \cap {\cal E}_B$ and there is no ellipsoid “sandwiched” between ${\cal E}_C$ and ${\cal E}_A \cap {\cal E}_B$. It is easy to see that that happens iff $\partial{\cal E}_C$ intersects $\partial{\cal E}_A$ by a $k$-dimensional ellipsoid and $\partial{\cal E}_C$ intersects $\partial{\cal E}_B$ by an $n-k$ dimensional ellipsoid. This can be stated in terms of matrices $A$, $B$, and $C$.
Consider a minimal $C$ s.t. $C \succeq A$ and $C \succeq B$. Then there are two subspaces $U$ and $V$ with ${\mathbb R}^n = U \oplus V$ such that $u^T Cu = u^T Au$ for $u\in U$ and $u^T Cu = u^T Bu$ for $u\in V$; in particular, $\operatorname{rank}(C-A) + \operatorname{rank}(C-B) \leq n$.
Visualization. The following animation shows ellipses ${\cal E}_C$ inscribed in ${\cal E}_A \cap {\cal E}_B$ in two dimensions. Every ellipse ${\cal E}_C$ corresponds to a minimal matrix $C$ ($C \succeq A$ and $C \succeq B$)
And this animation shows minimum ellipses ${\cal E}_C$ containing ${\cal E}_A \cup {\cal E}_B$.
I'm afraid that there is no more explicit characterization of sets $\{C: C \succeq A \text{ and } C \succeq B\}$ and $\{C: C \preceq A \text{ and } C \preceq B\}$.
Correspondence between “meet” and “join” matrices. Note that if $C_1$ is a “join” then $C_2 = A+B-C_1$ is a “meet” and vice versa. That follows from the fact that $C_2 \preceq A$ iff $A+B-C_1 \preceq A$ iff $B \preceq C_1$; similarly, $C_2 \preceq B$ iff $A \preceq C_1$.
Characterization for $2\times 2$ matrices. If $A$ and $B$ are $2\times 2$ matrices (s.t. $A\not\preceq B$ and $A\not\succeq B$) then meet and join matrices $C$ must satisfy the following equations $\det(C-A) = 0$ and $\det(C-A) =0$. The set of symmetric matrices that satisfy this system of equations forms a one dimensional curve in the space of matrices. Let us write
$$C_{xyz} = \begin{pmatrix}x&y\\y&z\end{pmatrix}.$$ Then the set
$$\{(x,y,z): \det(C_{xyz} - A) = \det(C_{xyz} - B) = 0\}$$
is a hyperbola (that lies in a plane in ${\mathbb R}^3$). Points on one branch of the hyperbola correspond to join matrices; points on the other branch correspond to meet matrices.
Notes. Note that by changing the basis we can always assume that $A=I$ and $B$ is a diagonal matrix, but I don't think that this observation leads to a very explicit characterization of $\{C: C \succeq A \text{ and } C \succeq B\}$. In particular, $C$ does not have to be a diagonal matrix. For example, let
$$A=\begin{pmatrix} 1& 0\\0& 1\end{pmatrix}\quad B =\begin{pmatrix} 2& 0\\0& 1/2\end{pmatrix}.$$ Then the following matrices $C$ are minimal matrices greater ($\succeq$) than $A$ and $B$:
$$C=\begin{pmatrix} 2& 0\\0& 1\end{pmatrix}\quad C=\begin{pmatrix} 3& 1\\1& 3/2\end{pmatrix} \quad C=\begin{pmatrix} 3& -1\\-1& 3/2\end{pmatrix}.$$
(The set of all such matrices $C$ forms a one dimensional curve in the space of all $2\times 2$ matrices.)
$S_+^n$ closed follows quite elementarily from definition, rather than by using topological properties of the eigenvalues of a matrix.
Let $x\in\Bbb R^n$ and consider the following linear maps:
$G:\Bbb R^{n\times n}\to \Bbb R^{n\times n}$, $G(A)=A-A^t$.
$q_x:\Bbb R^{n\times n}\to \Bbb R$, $q_x(A)=x^tAx$.
Since all these maps are continuous, $$S^n_+:=\ker G\cap \bigcap_{x\in\Bbb R^n} q_x^{-1}[0,\infty)$$ must be closed, because it's intersection of closed sets. As an additional observation, this is also an intersection of preimages of convex cones by linear maps, and thus a convex cone.
Best Answer
For closed, note that the functions $f_1:\mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$ given by $f_1(A) = A-A^T$, and $f_2: \mathbb{R}^{n\times n} \to \mathbb{R}$ given by $f_2(A) = \min_{||x||=1} \langle x, A x \rangle$ are continuous, so the sets $f_1^{-1} \{ 0 \}$ and $f_2^{-1} [0,\infty)$ are closed.
For convex, it is straightforward to check that if $\lambda \in [0,1]$, and $A,B \in S_n^+$, then $\lambda A + (1-\lambda)B \in S_n^+$ (just use the definition of positive semi-definiteness).
For cone, again it is straightforward to check that if $t\geq 0$ (0r whatever your definition of cone allows), then when $A \in S_n^+$, then $t A \in S_n^+$ (again, just use the definition).