Prove that $SEQ_{DFA}$ = {⟨A,B⟩ | A,B are DFAs and L(A) ⊆ L(B)} is decidable
There are several ways to prove that $SEQ_{DFA}$ is decidable. One of them is by using the fact that $EQ_{DFA}$ is decidable.

Assume that $L_1$ and $L_2$ are languages such that $L_1\leq_m L_2$ (that is, there is a mapping reduction from $L_1$ to $L_2$). It holds that if $L_2$ is decidable, then so is $L_1$. Hence,  in order to prove that $SEQ_{DFA}$ is deciadable, we reduce it to $EQ_{DFA}$. As hinted in the comment, we propose the following reduction.


The Reduction:
Let $C$ be a fixed DFA for the empty language, that is, $L(C) = \emptyset$. The reduction operates as follows. Given instance $\langle A= \langle Q_{A}, \Sigma, q^{A}_0, \delta_{A}, F_{A}\rangle, B= \langle Q_{B}, \Sigma, q^{B}_0, \delta_{B}, F_{B}\rangle\rangle$ of $SEQ_{DFA}$, the reduction computes a product automaton, $D$, for $L(A)\cap L(B)^C$ and outputs $\langle D, C\rangle$.
Correctness: $L(A)\subseteq L(B)$ iff $L(A)\cap L(B)^C = \emptyset$ iff $L(D) = \emptyset$ iff $L(D) = L(C)$. (the first equivalence mentioned in the comment is simple enough and thus is left for the reader).
Computability: the only non-trivial part is to compute $D$ given $A$ and $B$. We claim that $D$ is defined by $D = \langle Q_{A}\times Q_{B}, \Sigma, (q^{A}_0, q^{B}_0), \delta_{D}, F_{D}\rangle$. Where:


1) For every $(q, s)\in Q_{A}\times Q_{B}$ and $\sigma \in \Sigma$, $\delta_{D}$ is defined by: $$\delta_{D}((q, s),\sigma) = (\delta_{A}(q, \sigma), \delta_{B}(s, \sigma))$$ 

2) $F_{D} = F_{A}\times (Q_{B}\setminus F_{B})$.

Note that $D$ is a standard product automaton and thus for every finite word $w\in \Sigma^*$, we have that $\delta_{D}((q, s), w) = (\delta_{A}(q, w), \delta_{B}(s, w))$ (this can be proven by induction on $|w|$). Therefore, $w\in L(D)$ iff $\delta_{D}((q^{A}_0, q^{B}_0), w) \in F_{D}$ iff $(\delta_{A}(q^{A}_0, w),\delta_{B}(q^{B}_0, w) ) \in F_{A}\times (Q_{B}\setminus F_{B})$ iff $w\in L(A)$ and $w\in L(B)^C$ iff $w\in L(A)\cap L(B)^C$. Therefore $L(D) = L(A)\cap L(B)^C$ and thus we're done.