Consider an $n$ by $n$ matrix $A$, as arbitrary points on the $\mathbb{E}^{n^2}$ Euclidean space. Let $S$, $S_+$, and $S_{++}$ denote respectively the sets of symmetric, positive definite, and positive semidefinite $n$ by $n$ matrices on $\mathbb{E}^{n^2}$.
How to prove that $S$, $S_+$, and $S_{++}$ are convex sets?
Let's do it in $S$. Bear in mind that it's quite the same for $S_ +$ and $S_ {++}$.
In order to prove $S$'s convexity you only need to show that given any two matrices $A,B\in S$, the straight segment connecting $A$ and $B$ is completely contained in $S$.
Consider the path $$\gamma:[0,1]\to \mathbb E^{n^2};\quad t\mapsto \gamma(t) = (t-1)A + tB, $$ which is the straight segment connecting $A$ and $B$. We need to prove that, for all $t\in[0,1]$, $\gamma(t)\in S$. This is, we need to show that $\gamma(t)^T = \gamma(t)$, with $\cdot ^T$ denoting the transpose matrix. Note that $$\gamma(t)^T = ((t-1)A + tB)^T = (t-1)A^T + tB^T = (t-1)A + tB = \gamma(t),$$ since $A^T = A$ and $B^T = B$.
I bet you can replicate this process for $S_ +$ and $S_{++}$.