A vector function $F$ is monotonic in $S_i$ if $$ (F(x)-F(y))^T (x-y)\geq 0, \forall x,y\in S_i. $$
However, I am unsure if this property holds for continuous functions when considering the union of the sets, i.e., if a continuous vector function is monotonic in $S_1, S_2, \dots, S_n$ respectively, is it monotonic in the whole $\cup_{i=1}^n S_i$?
Without additional assumptions about the domain of $F$, the answer is no. For example, if $S_1 = \{(x,y) \in \mathbb R^2 ~;~ x < -1\}$, $S_2 = \{(x,y) \in \mathbb R^2 ~;~ 1 < x\}$ and $U = S_1 \cup S_2$, you can see that
$$F(x,y) = \begin{cases}(1,0) & \text{if } (x,y) \in S_1 \\ (-1,0) & \text{if } (x,y) \in S_2 \end{cases}$$
is continuous on $U$ and monotonic on $S_1$ and $S_2$. Now if you take $v \in S_1$ and $w \in S_2$, you have $F(w) - F(v) = (-2,0)$ and the first coordinate of the vector $w - v$ is positive so $F$ is not monotonic on $U$. Of course it works because $U$ is not a connected subset of $\mathbb R^2$.
If now $U \subset \mathbb R^n$ is a connected subset and $F : U \to \mathbb R^n$ is your continuous function, I have to think about it.
Edit : $S_1 = \{(x,y) \in \mathbb R^2 ; (x+1)^2 + y^2 \leq 1\}$, $S_2 = \{(x,y) \in \mathbb R^2 ; (x-1)^2 + y^2 \leq 1\}$, and $U$ is the union of these two circles. You can see that $U$ is path-connected. Define
$$F(x,y) = \begin{cases}(y,-x) & \text{if } (x,y) \in S_1 \\ (-y,x) & \text{if } (x,y) \in S_2 \end{cases}$$ $F$ is the rotation of angle $\frac{3\pi}{2}$ on $S_1$ while it is the rotation of angle $\frac{\pi}{2}$ on $S_2$ (both with center $(0,0)$). Now $F$ is well-defined, continous and monotonic on $S_1$ and $S_2$. Observe that $$\Big(F\big((1,1)\big) - F\big((-2,0)\big)\Big)^T \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = -4 < 0$$ Because $(1,1) \in S_2$ and $(-2,0) \in S_1$, we see that $F$ is not monotonic on $U$.
I’ll let you handle the case where $U$ is convex.
Edit: For convex U, we have:
If a vector-valued continuous function ${F}$ is monotonic in separated sets $U_i, i=1,\dots,n$ of a convex domain $U$ respectively, then it is monotonic in the whole domain $U$.
Proof. For any two points $x_1, x_K\in U$, by convexity of $U$, we suppose the segment between them (i.e., $\{\theta x_1 + (1-\theta)x_K| \theta\in [0,1]\}$) is crossed with boundaries of $U_k$ at $x_k = \theta_k x_1 + (1-\theta_k)x_K, k=2,\dots,K-1$ respectively. Without loss of generality, we consider $\theta_k < \theta_{k-1}$.
By the definition of monotonicity of vector-valued functions over $U_k$, we have $$ [F(x_k) - F(x_{k-1})]^T (x_k - x_{k-1}) \geq 0, $$ which means $$ [F(x_k) - F(x_{k-1})]^T (\theta_k - \theta_{k-1}) (x_1 - x_K)\geq 0. $$ Since $\theta_k < \theta_{k-1}$, we have $$ [F(x_k) - F(x_{k-1})]^T (x_1 - x_K)\leq 0. $$ By summing over $k$ from $2$ to $K$, we obtain $$ [F(x_K) - F(x_{1})]^T (x_1 - x_K)\leq 0, $$ which proves the monotonicity of $F$ over $U$.