Suppose $b$ is the smallest element of $B$. Then $b$ is also a minimal element of $B$. Working on the book: Daniel J. Velleman. "HOW TO PROVE IT: A Structured Approach, Second Edition" (p. 206)

Suppose $R$ is a partial order on a set $A$, and $B \subseteq A$.
  Suppose $b$ is the smallest element of $B$. Then $b$ is also a minimal element
  of $B$, and it is the only minimal element.

In this post, Suppose $b$ is the smallest element of $B$. Then $b$ is also a minimal element of $B$, and it is the only minimal element., I showed that the minimal element is unique. Now, I will prove premise two of that proof (that $b$ is also a minimal element of $B$).
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\oi#1{\qquad\mathbf{\lor I} \: #1 \\}
\def\oe#1{\qquad\mathbf{\lor E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\neg I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
$
\fitch{
1.\, \forall x\forall y((xRy \land yRx) \to x=y)\\
2.\, b \in B \land \forall x(x \in B \to bRx)
}{
 \fitch{3.\, \exists x(x \in B \land xRb \land x \neq b)}{
   \fitch{4.\, a \in B \land aRb \land a \neq b}{
     5.\,\forall x(x \in B \to bRx) \ce{2}
     6.\,a \in B \to bRa \Ae{5}
     7.\,a \in B \ce{4}
     8.\,bRa \ie{6,7}
     9.\,(aRb \land bRa) \to a = b \Ae{1}
     10.\,aRb \ce{4}
     11.\,aRb \land bRa \ci{10,8}
     12.\,a=b \ie{9,11}
     13.\,a \neq b \ce{4}
     14.\,\bot \ne{12,13}
  }\\
  15.\,\bot \Ee{3,4-14}
}\\
16.\,\neg \exists x(x \in B \land xRb \land x \neq b) \ni{3-15}
17.\,b \in B \ce{2}
18.\,b \in B \land \neg \exists x(x \in B \land xRb \land x \neq b) \ci{17,16}
}
$
Why do I need to add "$x \in B$" in the symbolisation of "$b$ is a minimal element of B" in order to accomplish the proof ?
Is this proof correct ?
 A: Your derivation is correct. Note that you only need that $R$ is antisymmetric to prove that the smallest element of $B$ is a minimal element of $B$.
The formula
$$\tag{1}\lnot \exists x (xRb \land x \neq b)$$
means that $b$ is a minimal element for the domain of quantification, that is the set $A$ and not the subset $B$. Indeed, $\lnot \exists x$ means that there is no $x$ in the domain.
The fact that $b$ is the smallest element of $B \subseteq A$ does not imply that $b$ is a minimal element of $A$. For instance, if $A = \{0,1\}$ and $B = \{1\}$ with the usual order, then $B \subseteq A$, and $1$ is the smallest element of $B$ but is not a minimal element of $A$.
This is the reason why it is important to add $x \in B$ to $(1)$ and get
$$\tag{2} \lnot \exists x (x \in B \land xRb \land x \neq b)$$
which actually means that there is no element in $B$ smaller than $b$. According to $(2)$, possibly $b \in A \smallsetminus B$, so formula $(2)$ alone does not mean that $b$ is a minimal element of $B$.
But formula $(2)$ in conjunction with $b \in B$ means that $b$ is a minimal element of $B$.
From a technical point of view, in your derivation you can easily see the need for adding $x \in B$ to $(1)$.
Suppose that your line $3$ were $\exists x (xRb \land x \neq b)$, which means that $b$ is not a minimal element of $A$.
As in line $3$ there would not be $x \in B$, you cannot infer $a \in B$ in line $7$ and hence you cannot derive $bRa$ in line $8$ (which is essential to use antisymmetry and conclude that $a = b$ in line 12, so that you get a contradiction and thus derive that $b$ is a minimal element of $B$).
