Can someone verify my direct proof that if $A$ is a subset of $B$, $A\cup B = B$? This is a problem from Discrete Mathematics and its Applications

I am trying to use a direct proof to do this problem. 
Here is my book's explanation/section on direct proof

Here is my work so far
I used this property of sets(from  https://courses.cs.washington.edu/courses/cse311/14au/slides/lecture09-filled.pdf)

Basically that two sets are equal if they are subsets of each other. 
I know that for a direct proof, I am trying to prove p -> q, or in this case, if A is a subset of B, then the union of A and B is equal to B.  My first step was assuming that  A is a subset of B meaning that if x is in A, x is also in B. Because I have an implication involving two propositions, I used the law of implication to convert p -> q into ~p v q.
My initial assumption ends up being X is not in A or x is in B.(assumption). I first tried showing that $A\cup B$ is a subset of B. To do this, I used the definition of $A\cup B$ to say that if X is a member of $A\cup B$, either X is in A or X is in B. From our initial assumption(X is not in A or X is in B), if X is in A, then X has to be in B(first condition isn't met, second has to be met). In the other case if X is in B, then X is in B. So In either case of the Union possibility, X ends up being a member of B. After this, I have shown that B is a subset of $A\cup B$.
My next step was to show that B is a subset of $A\cup B$. Or another way of saying it, if X is a member of B, then it is a member of $A\cup B$. By definition of union, if X is a member of $A\cup B$,  is either in A or in B. From that, my next logical step was saying that if X is a member of B, it is also a member of $A\cup B$  because it meets the second condition in the conjunction, in B.  I can now conclude that  $A\cup B$ is a subset of B.
Because I have shown that if A is a subset of B,  $A\cup B$  is a subset of B and B is a subset of $A\cup B$. By definition of equality of sets, $A\cup B$ is equal to B.
Does everything look logically coherent/organized? Is there any part I can rephrase to make it more understandable?
 A: Symbolically:
$\qquad A\cup B = B 
\\\equiv \Big\{ x\mid(x \in A\vee x\in B) \leftrightarrow x\in B \Big\}
\\ \equiv \Big\{ x \mid \big((x\in A\vee x\in B)\to x\in B\big) \wedge \big(x\in B\to (x\in B\vee x\in A)\big)\Big\}
\\ \equiv \Big\{ x \mid \big((x\in A\vee x\in B)\to x\in B\big) \wedge \top\Big\}
\\ \equiv \Big\{ x \mid \big((x\in A\vee x\in B)\to x\in B\big) \Big\}
\\ \equiv \Big\{ x \mid \big(x\in A\to x\in B\big)\wedge \big(x\in B \to x\in B\big)\Big\}
\\ \equiv \Big\{ x \mid \big(x\in A\to x\in B\big)\wedge \top\Big\}
\\ \equiv \Big\{ x \mid \big(x\in A\to x\in B\big)\Big\}
\\ \equiv A\subseteq B
$
Using this as a guideline we obtain a direct proof.


*

*Assuming $A \subseteq B$ then $A\cup B \subseteq B$ because every element in $A$ is in $B$ (from the assumption) and every element in $B$ is in $B$ (clearly), so every element in the union of the two must be in $B$.


*

*Additionally $A\cup B \supseteq B$ because every element in $B$ is in the union $A\cup B$ by definition.

*This means that $A\cup B = B$ when we assume that $A\subseteq B$.


*Assuming that $A\cup B = B$ then every element in $A$, being an element of the union, is an element of $B$, and hence we have $A\subseteq B$ when we assume $A\cup B = B$.
We conclude that $A\subseteq B$ is equivalent to $A\cup B = B$
A: I would write it something like this.

We want to show $A\cup B = B$.  There are two parts:

*

*First, we want to show $B\subseteq A\cup B$.  Assume $x\in B$. Then certainly $x\in A$ or $x\in B$, so we are done.


*Now we show $A\cup B\subseteq B$.  Assume $x\in A\cup B$; we want to show $x\in B$.  Since $x\in A\cup B$,  either $x\in A$ or $x\in B$. if  the latter case holds we are already done.  In the former case we can conclude that $x\in B$ because of the hypothesis that $A\subseteq B$.

I think this is not only easier to follow than a completely symbolic proof, and less prone to dumb errors, but it is also shorter.  (Also, it is easier for the grader to see that you understand what is going on.)
A: Looks reasonable. Your handwritten version doesn't exactly explain 
$$\begin{align}
x\in A \lor x \in B \\
\text{ and ... }x\not\in A \land x \in B \\
\text{ ... to ...}\\ 
x \in B\\
\end{align}$$
but you do explain that in the text here. I think I would have kept the implication $x\in A \implies x \in B $ and gone through like this:
$$\begin{align}
x\in A \lor x \in B \\
\implies x\in B \lor x \in B \\
\implies x \in B\\
\end{align}$$
and you could be briefer inferring that $B\subseteq A\cup B$.
