FALSE: a symmetric and transitive relation on a set $A$ is reflexive on $A$, so reflexivity is redundant in the definition of an equivalence relation

The argument that makes the claim look right: given $a\in A$, take any $b$ with $(a,b)\in R$; symmetry gives $(b,a)\in R$, and transitivity applied to $(a,b)$ and $(b,a)$ gives $(a,a)\in R$.

The witness: put $u:=\mathrm{\varnothing }$, $v:=\{\mathrm{\varnothing }\}$, $A:=\{u,v\}$ and $R:=\{(u,u)\}$. Here $u\mathrm{\ne }v$, because $v$ has an element and $u$ has none.

The gap in step 1.1 is the phrase "take any $b$ with $(a,b)\in R$": no hypothesis supplies such a $b$. Symmetry and transitivity are conditional on pairs that are already in $R$, so they constrain $R$ only at points that $R$ relates to something, and say nothing whatever about a point of $A$ that $R$ leaves untouched.

$R$ is a relation on $A$: its only element is the ordered pair $(u,u)$, whose coordinates both lie in $A$.

$R$ is symmetric, since its only pair is its own reversal, and transitive, since the only composable pair of its members is $(u,u)$ with $(u,u)$, whose conclusion $(u,u)\in R$ holds.

$R$ is not reflexive on $A$: $v$ is an element of $A$ and $(v,v)\mathrm{\notin }R$, since the only element of $R$ is $(u,u)$ and $v\mathrm{\ne }u$.

The claim is therefore false, and with it the conclusion drawn from it: the reflexivity clause in the definition of an equivalence relation is not redundant, since $R$ satisfies the other two clauses on $A$ and is not an equivalence relation on $A$.