How do we prove that $L \int_a^b f= L\int_a^c f + L \int\limits_c^b f$, where $L\int_a^b f$ is the lower integral of $f$ on $[a,b]$? In Spivak's Calculus, Chapter 14 "The Fundamental Theorem of Calculus", there is the following snippet

If $f$ is any bounded function on $[a,b]$, then
$$\sup\{L(f,P)\} \text{ and } \inf\{U(f,P)\}$$
will both exist, even if $f$ is not integrable. These numbers are
called the lower integral of $f$ on $[a,b]$ and the upper
integral of $f$ on $[a,b]$, respectively, and will be denoted by
$$L\int\limits_a^b f \text{ and } U\int\limits_a^b f$$
The lower and upper integrals both have several properties which the
integral possesses. In particular, if $a<c<b$, then
$$L \int\limits_a^b f= L\int\limits_a^c f + L \int\limits_c^b f\tag{1}$$
and
$$U \int\limits_a^b f= U\int\limits_a^c f + U \int\limits_c^b f$$
The proofs of these facts are left as an exercise, since they are
quite similar to the corresponding proofs for integrals. The results
for integrals are actually a corollary of the results for upper and
lower integrals, because $f$ is integrable precisely when
$$L\int\limits_a^b f = U\int\limits_a^b f$$

How do we prove $(1)$?
Here is my attempt

Let
$$P_1=\{a=t_0,t_1,...,t_{m-1},c=t_m\}$$
$$P_2=\{ c=u_0, u_1,...,u_{n-1},b=u_n \}$$
$$P=\{ a=t_0,t_1,...,t_{m-1}, c=t_m=u_0, u_1,...,u_{n-1},b=u_n\}$$
be partitions of $[a,b]$. Then $P$ contains $P_1$ and $P_2$.
Then
$$L(f,P)= \sum\limits_{i=1}^m
 m_i\Delta t_i + \sum\limits_{i=1}^n m_i \Delta u_i$$
$$L\int\limits_a^b f = \sup\{L(f,P)\}= \sup\left \{
 \sum\limits_{i=1}^m
 m_i\Delta t_i + \sum\limits_{i=1}^n m_i \Delta u_i \right  \}$$
$$=\sup\left \{ \sum\limits_{i=1}^m m_i\Delta t_i \right \}+\sup\left
 \{ \sum\limits_{i=1}^n m_i \Delta u_i \right  \}$$
$$=\sup\{ L(f,P_1)\}+\sup\{ L(f,P_2)\}$$
$$L\int\limits_a^c f + L\int\limits_c^b f$$
$$\blacksquare$$

Is this correct?
As an additional detail, let's see how we obtain the corresponding result for integrals?
Note that $(1)$ is true for any bounded function $f$. $f$ doesn't need to be integrable. However, if if is integrable then
$$L\int\limits_a^b f = U\int\limits_a^b f=\int\limits_a^b f$$
Then all we have to do is sub into $(1)$. That is, since
$$L \int\limits_a^b f= L\int\limits_a^c f + L \int\limits_c^b f$$
Then
$$\int\limits_a^b f= \int\limits_a^c f + \int\limits_c^b f$$
 A: Correct Proof
Using arbitrary partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$, we form $P = P_1 \cup P_2$ which is a partition of $[a,b]$ and we have
$$L(f,P_1) + L(f,P_2) = L(f,P) \leqslant L\int_a^b f$$
The partitions $P_1$ and $P_2$ can be varied independently and it follows by taking suprema over $P_1$ and $P_2$ succesively that
$$\tag{1}L\int_a^c f + L\int_c^b f \leqslant L \int _a^b f$$
To prove the reverse inequality, note that for any $\epsilon > 0$ there exists a partition $P$ of $[a,b]$ that may not include the point $c$ such that
$$L\int_a^b f - \epsilon < L(f,P)$$
Adding the point $c$ to $P$ if necessary, we get the partition P' which is a refinement of $P$ and since lower sums are increasing under partition refinement it follows that
$$L\int_a^b f - \epsilon < L(f,P) \leqslant L(f,P')$$
Now $P'$ which includes $c$ is the union of partitions  $P_1'$ of $[a,c]$ and $P_2'$ of $[c,b]$. Hence,
$$L\int_a^b f - \epsilon < L(f,P') = L(f,P_1') + L(f,P_2') \leqslant L\int_a^cf + L\int_c^b f$$
Since this is true for any $\epsilon > 0$ it follows that
$$\tag{2} L \int_a^b f \leqslant L \int_a^c f + L \int _c^b f$$
Together (1) and (2) imply that
$$ L \int_a^b f = L \int_a^c f + L \int _c^b f$$
