If $f$ is an integrable function on $[a,b]$, then $f$ is integrable on $[c,d]$ $\forall [c,d]\subseteq [a,b]$ 
If $f$ is an integrable function on $[a,b]$, then $f$ is integrable on $[c,d]$ $\forall [c,d]\subseteq [a,b]$

My solution:
Let $P'=\{a,c,d,b\}$ be a partition of $[a,b]$. Let $P$ be a refinement partition of $P$ such that $||P||<\delta$
$f$ is integrable on $[a,b]\implies \exists P\text{ of } [a,b] \text{ s.t. }\forall \epsilon>0~ S(P)-s(P)\leq \epsilon $ where $S(P),s(P)$ is the upper and lower Riemann sums over the partion P. 
$$\sum^n_{i=0}\left(M_i-m_i\right)\left(x_i-x_{i-1}\right)\leq \epsilon$$
Suppose at $i=k$ we have $(c,x_{k-1})$ and at $i=k+m$ we have $(x_{k+m},d)$, Since each term in the summation is nonnegative then:
$$\sum^{k+m}_{i=k}\left(M_i-m_i\right)\left(x_i-x_{i-1}\right)\leq \sum^{k}_{i=0}\left(M_i-m_i\right)\left(x_i-x_{i-1}\right)+\sum^{k+m}_{i=k}\left(M_i-m_i\right)\left(x_i-x_{i-1}\right)+\sum_{k+m+1}^{n}\left(M_i-m_i\right)\left(x_i-x_{i-1}\right)\leq \epsilon$$
$$\implies\text{ f is integrable on }[c,d]$$
Is my solution correct?
 A: You wanting to prove that $f$ is integrable of $[c,d]$, i.e.

For every $\epsilon>0$ there exists $\delta>0$ such that for  every partition $P$ of $[c,d]$ that is finer than $\delta$, we have $S(P)-s(P)<\epsilon$.

So start with "Let $\epsilon>0$ be given." Then use integrability of $f$ on $[a,b]$ to obtain some suitable $\delta$ (that may depend on $\epsilon$). Then assume that $P$ is any partition of $[c,d]$ that is finer than $\delta$.  From this, construct a suitable partition $P'$ of $[a,b]$ that allows you to bound $S(P)-s(P)$ (presumably in terms of $S(P')-s(P')$).
A: No, it is not correct. First, there are these symbols that you introduce ($\delta$, $\varepsilon$) without saying what they are. Then you use the letter $P$ for two distinct things.
Let $\varepsilon>0$. You want to prove that there is a partition $P$ of $[c,d]$ such that, if $P=\{a_0,a_1,\ldots,a_n\}$, with $c=a_0<a_1<\cdots<a_n=d$, and if, for each $i\in\{1,2,\ldots,n\}$, $M_i=\sup f|_{[a_{i-1},a_i]}$ and $m_i=\inf f|_{[a_{i-1},a_i]}$, then$$\sum_{i=1}^n(M_i-m_i)(a_i-a_{i-1})<\varepsilon.\tag{1}$$Let $Q$ be a partition of of $[a,b]$ such that, if $Q=\{a_0,a_1,\ldots,a_n\}$, with $a=a_0<a_1<\cdots<a_n=b$, and if, for each $i\in\{1,2,\ldots,n\}$, $M_i=\sup f|_{[a_{i-1},a_i]}$ and $m_i=\inf f|_{[a_{i-1},a_i]}$, then you have $(1)$. Then there are $k,l\in\{1,2,\ldots,n-1\}$ such that $k<l$, that $c\in[a_k,a_{k+1}]$, and that $d\in[a_l,a_{l+1}]$. Consider the partition $P=\{c,a_{k+1},a_{k+2},\ldots,a_l,d\}$ of $[c,d]$. Then you have $(1)$.
