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Presumably there's a typo in the question, otherwise the relation is neither antisymmetric nor transitive. Counterexamples:
E.g. $\big((1,2),(2,3)\big) \in R$ since $1<2$, and $\big((2,3),(1,2)\big) \in R$ since $2<3$, and thus $R$ is not antisymmetric. E.g. $\big((1,1),(1,2)\big) \in R$ since $1=1$ and $1 \leq 2$, and $\big((1,2),(3,2)\big) \in R$ since $1<2$, but $\big((1,1),(3,2)\big) \not\in R$ since $1 \not< 1$ and $3 \not\leq 2$, and thus $R$ is not transitive.
This might explain why you are finding the question difficult.
Fixing the probable typo, the relation becomes:
$\big((x_1,x_2),(y_1,y_2)\big) \in R$ if and only if $x_1<y_1$ or $x_1=y_1$ and $x_2 \leq y_2$.
To show it's an order relation, we check:
Reflexive: Since $x_1=x_1$ and $x_2 \leq x_2$, we know $\big((x_1,x_2),(x_1,x_2)\big) \in R$. Hence $R$ is reflexive. Antisymmetric: Suppose $\big((x_1,x_2),(y_1,y_2)\big), \big((y_1,y_2),(x_1,x_2)\big) \in R$. Then there are four possibilities:
$$
\begin{array}{c|c|c|}
& x_1<y_1 & x_1=y_1 \text{ and } x_2 \leq y_2 \\
\hline
y_1<x_1 & \text{impossible} & \text{impossible} \\
\hline
y_1=x_1 \text{ and } y_2 \leq x_2 & \text{impossible} & \\
\hline
\end{array}
$$
So, we must have $x_1=y_1$ and $x_2 \leq y_2$ and $y_2 \leq x_2$. Since $\leq$ is an antisymmetric (order) relation, we must have that $x_2=y_2$. Hence $(x_1,x_2)=(y_1,y_2)$, and we can conclude that $R$ is antisymmetric. Transitive: Suppose $\big((x_1,x_2),(y_1,y_2)\big), \big((y_1,y_2),(z_1,z_2)\big) \in R$. Then there are four possibilities:
$$
\begin{array}{c|c|c|}
& x_1<y_1 & x_1=y_1 \text{ and } x_2 \leq y_2 \\
\hline
y_1<z_1 & \text{implies } x_1<z_1 & \text{implies } x_1<z_1 \\
\hline
y_1=z_1 \text{ and } y_2 \leq z_2 & \text{implies } x_1<z_1 & \text{implies } x_1=z_1 \text{ and } x_2 \leq z_2 \\
\hline
\end{array}
$$
(The above deductions use the transitivity of the relations $=$, $<$ and $\leq$, along with the properties "$x_1=y_1<z_1$ implies $x_1<z_1$" and "$x_1<y_1=z_1$ implies $x_1<z_1$".) In all four cases, $\big((x_1,x_2),(z_1,z_2)\big)$ satisfies the conditions of $R$. Hence $\big((x_1,x_2),(z_1,z_2)\big) \in R$, and we can conclude that $R$ is transitive.
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