The RTT Yangian has its standard Hopf structure

Each coefficient of $T^{[1]}(u)T^{[2]}(u)$ is a finite sum: the coefficient of $u^{-n}$ uses only the finitely many pairs of exponents adding to $n$ and the finite index sum $1\le k\le N$. Thus the proposed coproduct lands in the ordinary algebraic tensor product. For two independent Yangian factors, write $T_s^{[r]}$ for the matrix acting on auxiliary leg $s\in \{1,2\}$ and coefficient factor $r\in \{1,2\}$. Then $T_1^{[2]}$ commutes with $T_2^{[1]}$, and $T_1^{[1]}$ commutes with $T_2^{[2]}$, because each pair acts on disjoint tensor factors. Reordering these pairs and applying the RTT relation first in coefficient factor $1$ and then in factor $2$ gives $R_{12}{T}_1^{[1]}{T}_1^{[2]}{T}_2^{[1]}{T}_2^{[2]}=T_2^{[1]}{T}_2^{[2]}{T}_1^{[1]}{T}_1^{[2]}{R}_{12}$. The middle matrix product is precisely the RTT relation for $\mathrm{\Delta }(T)$, so $\mathrm{\Delta }$ preserves the defining relation and extends to a unital algebra homomorphism.

Iterating the matrix product yields $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(T)=T^{[1]}{T}^{[2]}{T}^{[3]}=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(T)$, so the coproduct is coassociative coefficientwise. Substitution $T\mapsto I$ sends the RTT relation to $R=R$, so it defines a unital counit homomorphism; it also gives $(ϵ\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(T)=T=(\mathrm{i}\mathrm{d}\otimes ϵ)\mathrm{\Delta }(T)$ and $ϵ(1)=1$.

Since $T(u)=I+O(u^{-1})$, recursively solving $T(u)T(u{)}^{-1}=I$ gives a unique right inverse; solving in the reverse order gives a left inverse, and associativity makes them equal. Each inverse coefficient is a finite polynomial in the coefficients of $T-I$, because only products of at most $n$ positive-degree terms contribute to $u^{-n}$. Write $A=T_1(u)$ and $B=T_2(v)$. In $RAB=BAR$, multiply on the left by $A^{-1}{B}^{-1}=(BA{)}^{-1}$ and on the right by $B^{-1}{A}^{-1}=(AB{)}^{-1}$ to obtain $A^{-1}{B}^{-1}R=R{B}^{-1}{A}^{-1}$. Define an anti-homomorphism on the free generator algebra by sending the coefficient matrix $T(u)$ entrywise to this inverse matrix. Applying it to the defining RTT relator reverses coefficient multiplication, so its image is $R{B}^{-1}{A}^{-1}=A^{-1}{B}^{-1}R$, exactly the same identity. Hence it preserves the RTT ideal and descends to the quotient, with $S(T(u))=T(u{)}^{-1}$.

The two matrix inverse identities give ${\sum }_{k}S(t_{ik}(u))t_{kj}(u)={\delta }_{ij}={\sum }_k{t}_{ik}(u)S(t_{kj}(u))$, the left and right antipode identities on generators. For the left identity on a product $xy$, use $\mathrm{\Delta }(xy)=\sum x_{(1)}{y}_{(1)}\otimes x_{(2)}{y}_{(2)}$ and anti-multiplicativity of $S$ to obtain $\sum S(y_{(1)})S(x_{(1)})x_{(2)}{y}_{(2)}=ϵ(x)\sum S(y_{(1)})y_{(2)}=ϵ(x)ϵ(y)1$; the right identity follows from $\sum x_{(1)}{y}_{(1)}S(y_{(2)})S(x_{(2)})$. Induction on word length and linearity extend both identities to the generated algebra, giving $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }=\eta ϵ=m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }$.