$τ = \{U ⊆ E : A ⊆ U\} ∪ \{∅\}$. Prove that $τ$ is a topology, and then compute $cl(B)$ and $int(B)$ for all $B ⊆ E$. Let $E$ be a set and $A ⊆ E$. Define:
$τ = \{U ⊆ E : A ⊆ U\} ∪ \{∅\}$.
Prove that $τ$ is a topology, and then compute $cl(B)$ and $int(B)$ for all $B ⊆ E$.
1) proving it is a topology:
i) $\emptyset \in \tau$ by the above definition. Also, $E \subseteq E, A \subseteq E \implies E \in \tau$
ii) Let $U,V \in \tau \implies A\subseteq U \wedge A\subseteq V \implies A\subseteq U \cap V$ 
iii) let $\mathscr{U} \subseteq \tau \implies \forall U \in \mathscr{U},\,\, A \subseteq U \implies A \subseteq \displaystyle \bigcup \mathscr{U} \implies \displaystyle \bigcup \mathscr{U} \in \tau$
2) Computing the closures and interiors. This is the part I need help with. 
For all $B \subseteq E$, s.t. $A \subseteq B,\,\, int(B) \in \tau$
if $A \not\subseteq B \implies int(B) \not\in \tau$ 
$cl(B) \not\in \tau$ because it's closed so I am not really sure what there is to compute for these.
 A: First detail: we should write $\tau = \{ U\subseteq E \mid A \subseteq U \}\cup \color{red}{\{}\varnothing\color{\red}{\}}$ instead of what is there. Your verification that $\tau$ is a topology is fine, good job.
Now let $B \subseteq E$.
For computing ${\rm int}(B)$, we have two options: or $A \subseteq B$ or $A \not\subseteq B$. In the first case $B$ is open and so ${\rm int}(B)=B$, and in the second case, there is no open set contained in $B$, so ${\rm int}(B) = \varnothing$ (otherwise $U \subseteq B$ being open implies $A \subseteq B$ again).
Now we go for ${\rm cl}(B)$. Again, two options: or $A \cap B = \varnothing$ or $A \cap B \neq \varnothing$. In the first case, $A \subseteq (E \setminus B)$, so $E \setminus B$ is open, hence $B$ is closed and ${\rm cl}(B) = B$. The second case is tricky, but we can prove that ${\rm cl}(B) = E$ as follows: let $F\subseteq E$ be a closed set containing $B$ - let's check that $F = E$. We have: $$F \supseteq B \implies E \setminus F \subseteq E \setminus B.$$Now $E \setminus F$ is open, so or it contains $A$ or it is empty. But $A \subseteq E \setminus F$ leads to $A \subseteq E \setminus B$, contradicting $A \cap B \neq \varnothing$. So $E \setminus F = \varnothing$ and $F= E$. We conclude that ${\rm cl}(B) = E$, as wanted.
Bear in mind all the time that $${\rm int}(B) = \bigcup\{U \subseteq B \mid U \in \tau \}\,\quad {\rm cl}(B) = \bigcap\{F \supseteq B \mid E \setminus F \in \tau\}.$$
