Prove that every finite non empty set has an smallest element Suppose R is a total order on a set A.
I am done with base case. Assuming that any subset B (having n elements) of A has smallest element.
Assume $B'=B-{b}$. Now B' has n elements, so it has smallest element say c. $c \in B'$
Case1: $bRc$
Claim : $b$ is smallest element of B
If not, then  $\exists x \in B , xRb $. So $xRc$ for some $x \in B$ by transitivity. That $x =b$ because $\forall x \in B',   cRx $. So $cRb$ which is a contradiction to our assumption of $bRc$ as R is total order on A
Case 2: $\lnot bRc $
Claim : c is smallest element of B
If not, 
$\exists x \in B, xRc$. Also $\forall x \in B',   cRx $. So that $x=b so bRc$ which is a contradiction
Is my proof correct?
 A: Your proof looks correct to me but the beginning is a little cluttered for my taste, you should state more explicitely what you are trying to proof, like:
"We assume that every subset of $A$ with $n$ elements has a smallest element.
We will show that then every subset of $A$ with $n+1$ elements has a smallest element as well.
Let $B$ be a subset of $A$ with $n+1$ elements ..." [And so on, the rest of your proof is more technical than required but correct and clear.]
Also note William Elliot's comment on your notation.

Edit: I should be a little stricter with your style of writing proofs. When I wrote above the rest of your proof was "correct and clear" I failed to point out that there was still room for improvement.
I tried my best to rewrite your proof in a way that I personally find more appealing. This is POV but I invite you to read my version and compare the differences. I use a different notation for the relation / total order here and for the set difference, but your notation is equally fine except for the fact that it must be $B-\left\{ b\right\}$ (braces!) as $b$ without braces is not a set.
Definition: Let $B$
  be a set with total order $R\subset B\times B$. An element $s\in B$
  is called a "smallest element" with respect to $R$, if there is no other element $t\in B$, $t\neq s$ such that $\left(t,\, s\right)\in R$.
Assertion: Let $B$ be a finite non-empty set, with $N=\left|B\right|$ elements equipped with a total order $R\subset B\times B$. Then $B$
  has a smallest element.
Proof:
Let $N=1$, $B=\left\{ b\right\}$. Since there is no other element in $B$, $b$ is a smallest element.
For arbitrary $N\in\mathbb{N}$ assume that any finite non-empty set with at most $N-1$ elements has a smallest element.
Let $B^{\prime}\;:=\; B\backslash\left\{ b\right\}$. As $\left|B^{\prime}\right|=N-1$, there is a smallest element $c\in B^{\prime}$.
Case 1: $\left(b,c\right)\in R$.
Claim: $b$ is a smallest element of $B$.
If not, there exists an $x\in B\backslash\left\{ b\right\} =B^{\prime}$ such that $\left(x,b\right)\in R$. Since $c$ was a smallest element of $B^{\prime}$ and $R$ ist a total order it follows that $\left(c,x\right)\in R$ and by transitivity $\left(c,b\right)\in R$. Thus $b=c$ but $b\notin B^{\prime}$ while $c\in B^{\prime}$! Contradiction!
Case 2: $\left(c,b\right)\in R$.
Claim: $c$ is a smallest element of $B$.
If not, there exists an $x\in B$ such that $\left(x,c\right)\in R$. Since $c$ is a smallest element of $B^{\prime}$ we have $x\notin B^{\prime}$ and since $B\backslash B^{\prime}=\left\{ b\right\}$ we have $x=b$. This however means $\left(b,c\right)\in R$ and thus again $b=c$ resulting in the same contradiction as in case 1.
qed
Remark: I use the term "a smallest element" instead of "the smallest element" as I ommited the - albeit trivial - proof that a smallest element is unique.
Edit:
As William Elliot commented, there is a shorter straight forward proof for 2:
We show for all $x\in B$ that $\left(x,c\right)\notin R$. As $c$ is a smallest element of $B^{\prime}$ this is true if $x\in B^{\prime}$. Also $\left(b,c\right)\notin R$ since $b\neq c$. As $B\backslash B^{\prime}=\left\{ b\right\}$ that completes the proof.
