Let A and B be sets. Show that A is a subset of B if and only if for any set C, one has A union C is a subset of B union C. Can you verify my proof if it is right?
Let $A$ and $B$ be sets. 
(a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$.
(b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.
Edited Solution :
(a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.
     Solution 1 :(<=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.
     Solution 2 :(<=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.
(b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.
(<=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B.
 A: Let me analyze your proof attempt, and provide some hints for correction.

What i did for (a) was :
Since A is a subset of B, (=>)
x is a member of A , x is a member of B. Therefore, since x is a member of A, x is a member of A union C as well. Since x is a member of B as well, x is a member of B union C. I conclude that A union C is a subset of B union C.

What you've done here is to show that an arbitrary element of $A$ must also be an element of $B\cup C.$ This is true, but it only shows that if $A\subseteq B,$ then for any set $C$ we have $A\subseteq B\cup C.$ This isn't what you want to show, or at least, isn't all of it. Instead, start with some $x\in A\cup C,$ and show that it must be in $B\cup C.$ If $x\in A,$ then by your reasoning, we have $x\in B\cup C,$ but what if $x\notin A$?

Since A union C is a subset of B union C, (<=)
For A union C , x is a member of A or x is a member of C.
Take x is a member of A. A is a subset of A union C. (By definition of union)
For B union C, x is a member of B or x is a member of C.
Take x is a member of B. B is a subset of B union C. (By definition of union)
Since A union C is a subset of B union C , i conclude that A is a subset of B.

You want to take an arbitrary $x\in A,$ and show that $x\in B.$ Now, it's true that we can conclude that $x\in A\cup C$ for every set $C,$ and so $x\in B\cup C$ for every set $C.$ But why does this let us conclude that $x\in B$? Can you think of any particular sets $C$ that let us make this leap?

What i did for part (b) was :
Since A is a subset of B, (=>)
x is a member of A , x is a member of B. Therefore since x is a member of A, x is a member of A intersect C. Since x is a member of B as well, x is a member of B intersect C. Therefore, i conclude that A intersect C is a subset of B intersect C.

Your first mistake in reasoning happens when you conclude that since $x\in A,$ then $x\in A\cap C.$ This need not be true for every set $C.$ In fact, there are many sets $C$ for which it isn't true. Likewise, in general, an element of $B$ need not be an element of $B\cap C.$ More than that, though, you're starting from the wrong place. You want to show that $A\cap C\subseteq B\cap C$ for every set $C.$ To do this, you should start with an arbitrary set $C,$ and an arbitrary element $x$ of $A\cap C,$ then show that $x\in B\cap C.$ This is pretty straightforward. See what you can do.

Since A intersect C is a subset of B intersect C, (<=)
For A intersect C, x is a member of A and x is a member of C as well.
  Therefore A intersect C is a subset of A. (By definition of intersection)
  For B intersect C, x is a member of B and x is a member of C as well.
  Therefore B intersect C is a subset of B. (By definition of intersection)
Since A intersect C is a subset of B intersect C, i conclude that A is a subset of B.

So far, you've noted that $A\cap C\subseteq A$ and $B\cap C\subseteq B$ for any set $C.$ While true, this isn't helpful. You'd like to show that $A\subseteq B,$ which means we need to start with some $x\in A$ and show that $x\in B.$ But all you know is that for any set $C,$ we have $A\cap C\subseteq B\cap C.$ This part may be the trickiest. The idea here is to find a set $C$ such that $A\cap C=A$ and $B\cap C=B.$ Once you've found such a set, the rest is very straightforward. Try using Venn diagrams to figure out a set $C$ that works. Do you know how to use Venn diagrams to find intersections and unions?

Added:
Your second attempt shows more promise, but there are still some issues.

(a)  (=>) Given A is a subset of B, this implies A U B = B.

True, but why? Have you already proved this?

Given a set C, A U B U C = B U C.

Again, this is true, but why?

Since A is a subset of B, this implies A U C is a subset of B U C.

But this is what you should be proving, and you haven't actually finished the job. You're on your way, though. Try to justify the following inclusions and equalities: $$A\cup C\subseteq (A\cup C)\cup B=A\cup B\cup C=B\cup C.$$

Solution 1 :(<=) Given A U C is a subset of B U C. For A U C, x is a member of A U C . For B U C, x is a member of B U C.

I'm not entirely sure what the last two sentences are supposed to mean, here. Do you mean that you choose $x$ to be a member of $A\cup C,$ from which it follows by the given condition that $x\in B\cup C$? Remember, though, that you're trying to show that $A\subseteq B.$ So, instead, you should let $x$ be an arbitrary member of $A,$ and try to show that $x\in B$ by choosing $C$ carefully.

Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.

Here, you have a good setup for a proof by contradiction, but you don't actually follow through with it. Your contradictory conclusion doesn't actually follow. Consider $A=C=\{1\},$ $B=\emptyset.$ That's an instance where we can have $A\nsubseteq B,$ but still have $A\cup C\subseteq B\cup C.$ Your second approach shows more promise.

Solution 2 :(<=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C.

Again, not sure what this means.

Suppose C is a disjoint set.

You may have the right idea here. Disjoint from what?

Since A U C is a subset of B U C, it follows that A is a subset of B.

Depending on what you mean above, this may well work. Still, it's a good idea to actually justify it, rather than just saying "it follows." Textbooks will often just say "it follows," and leave the details to the reader, but it's a bad personal practice.

(b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A.

You're off to a great start!

Let x be a member of B and member of C.

Oh, no! This is what you're trying to show. You can't just declare it. However, you're on your way, if somewhat awkwardly. Instead, start with some $x\in A\cap C.$ By definition of intersection, we have $x\in A$ and $x\in C.$ Since $x\in A$ and $A\subseteq B,$ then $x\in B.$ Since $x\in B$ and $x\in C,$ then $x\in B\cap C.$ Since $x$ was an arbitrary member of $A\cap C,$ then we have $A\cap C\subseteq B\cap C,$ and since $C$ was an arbitrary set, then this is true for any set $C.$

(<=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well.

You've almost got the right idea, here. Instead, let $C$ be a superset of $B$ and a superset of $A$, and see if you can take it from there. Can you think of any examples of such a set $C$?

C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well.

Both of these are very true.

Since A intersect C is equals to B intersect C, A is a subset of B.

Uh oh! This doesn't follow. Consider $A=\{1\},$ $B=\{2\},$ and $C=\emptyset.$ We can certainly see that $A\cap C=B\cap C=C,$ but we don't have $A\subseteq B.$
A: Just for comparison, here is a different style proof that I find simpler to design and simpler to follow.  You are asked to prove that for any $\;A,B\;$,
$$
A \subseteq B \;\equiv\; \langle \forall C :: A \cup C \subseteq B \cup C \rangle
$$

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
$Starting at the most complex side, expanding the definitions, and simplifying gives us, for any $\;A,B\;$,
$$\calc
\langle \forall C :: A \cup C \;\subseteq\; B \cup C \rangle
\calcop\equiv{definitions of $\;\subseteq\;$ and of $\;\cup\;$ twice}
\langle \forall C, x :: x \in A \lor x \in C \;\Rightarrow\; x \in B \lor x \in C \rangle
\calcop\equiv{logic: $\;P \Rightarrow Q\;$ to $\;\lnot P \lor Q\;$ for easier manipulation; DeMorgan}
\langle \forall C, x :: (x \not\in A \land x \not\in C) \lor x \in B \lor x \in C \rangle
\calcop\equiv{logic: simplify using negation of $\;x \in C\;$ on other side of $\;\lor\;$}
\langle \forall C, x :: x \not\in A \lor x \in B \lor x \in C \rangle
\calcop\equiv{logic: move $\;\forall C\;$ to the only place where it is actually used}
\langle \forall x :: x \not\in A \lor x \in B \lor \langle \forall C :: x \in C \rangle \rangle
\calcop{\tag{*}\equiv}{set theory: no $\;x\;$ is in all $\;C\;$: take $\;C := \emptyset\;$; simplify}
\langle \forall x :: x \not\in A \lor x \in B\rangle
\calcop\equiv{logic: write $\;\lnot P \lor Q\;$ as $\;P \Rightarrow Q\;$; definition of $\;\subseteq\;$}
A \subseteq B
\endcalc$$

The crucial step $\Tag{*}$ here is that $\;\langle \forall C :: x \in C \rangle\;$ is false, since it is false for $\;C := \emptyset\;$.
Note how there was no need for proving both directions separately.
A: For the second let x be in A , since for any set C , AUC is subset of BUC we can take C as the complement of  the single element set containing  x then it follows that x must be in B 
