This is not true even for unions of two compact convex sets.

To construct a suitable counterexample, consider the unit disk $$D=\{z\in\mathbb C:|z|\le 1\}$$in the complex plane $\mathbb C$. By $\partial D:=\{z\in\mathbb C:|Z|=1\}$ we denote the boundary circle of the disk $D$.

Next, choose a decreasing sequence of real numbers $(r_n)_{n\in\omega}$ such that $\lim_{n\to\infty}r_n=\inf_{n\in\omega}r_n=1$ and the points $z_n=r_ne^{i\pi/2^n}$ do not see each other behind the disk, which means that for any distinct numbers $n,m$ the interval $[z_n,z_m]:=\{tz_n+(1-t)z_m:t\in[0,1]\}$ intersects the disk $D$. Let $C$ be the (closed) convex hull of the compact set $D\cup\{z_n\}_{n\in\omega}$. It is easy to see that $C\setminus D$ has infinitely many connected components (homeomorphic to the triangle with a removed side).

Now consider the compact convex sets $A_1=C\times\{0\}$ and $A_2=D\times[-1,1]$ in $\mathbb C\times\mathbb R\cong \mathbb R^3$.

It can be shown that the union $A:=A_1\cup A_2$ is not homeomorphic to a simplicial complex (even to a CW-complex).

Assuming that $A$ is homeomorphic to a CW-complex, we can use the domain invariance theorem to show that the boundary
$$\partial A=((C\setminus D)\times\{0\})\cup (\partial D\times [-1,1])\cup (D\times\{-1,1\})$$
of $A$ in $\mathbb C\times\mathbb R$ is contained in the 2-skeleton of the CW-complex. Using the domain invariance theorem once more, it can be shown that $\partial C\cup\partial D$ is contained in the 1-skeleton of the CW-complex and hence it is homeomorphic to a finite graph, which is not true.

**Conclusion.**

The union of finitely many convex sets in $\mathbb R$ is homeomorphic to a simplicial complex.

For $n\ge 3$ the union of two compact convex set in $\mathbb R^n$ can be non-homeomorphic to a CW-complex.

**Problem.** What is the situation in dimension 2? Is the union of finitely many compact convex sets in the plane homeomorphic to a simplicial complex?