4 the editings overlapped, apparently, it should be fixed now, sorry...

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that $(X,<_1)$ is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and $(X,<_1)$. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to $<_2$) which, endowed with the order $<_2$, is isomorphic to $\kappa$. For $\alpha<k$ define $$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha\; \beta> y_{\alpha'}\textrm{\ and\ }\beta>_2 y_{\alpha'}\}.$$ The set above is not empty, and so $y_\alpha$ is well defined, because $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}\}|y_{\alpha'}\}|<k$ and also $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|y_{\alpha'}\}|<k$ (this depends on $\{y_{\alpha'}\}_\{alpha'\{y_{\alpha'}\}_{\alpha'<\alpha}$ not being cofinal in $(k,<)$ and $(X',<_2)$), so their complements in $X'$ intersect. The set $Y=\{y_\alpha\}_{\alpha<k}$ is what we were looking for.

3 fixed some of the LaTeX by adding backticks, but gave up on one part

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that $(X,<_1)$ is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and $(X,<_1)$. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to $<_2$) which, endowed with the order $<_2$, is isomorphic to $\kappa$. For $\alpha<k$ define $$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha,\; <\alpha\; \beta> y_{\alpha'}\textrm{ y_{\alpha'}\textrm{\ and\ }\beta>_2 y_{\alpha'}\}.$$ The set above is not empty, and so $y_\alpha$ is well defined, because $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}\}|<k$ and also $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|<k$ (this depends on $\{y_{\alpha'}\}_{\alpha'\{y_{\alpha'}\}_\{alpha'<\alpha}$ not being cofinal in $(k,<)$ and $(X',<_2)$), so their complements in $X'$ intersect. The set $Y=\{y_\alpha\}_{\alpha<k}$ is what we were looking for.

2 fixed formatting problems

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that $(X,<_1)$ is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and $(X,<_1)$. <_1)$. Let us now construct$Y$using transfinite induction. Let$X'\subseteq X$be the initial segment of$X$(with respect to $<_2$) <_2$) which, endowed with the order $<2$<_2$, is isomorphic to$\kappa$. For $\alpha\alpha=\min{\beta\in \alpha<k$ define $$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha\; beta> y_{\alpha'}\textrm{<\alpha,\; \beta> y_{\alpha'}\textrm{ and \ }\beta>2 y{\alpha'}}.$$ \beta>_2 y_{\alpha'}\}.$$ The set above is not empty, and so$y_\alpha$is well defined, because $|{\gamma\in |\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}}|2)$)y_{\alpha}\}|<k$ and also $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|<k$ (this depends on $\{y_{\alpha'}\}_{\alpha'<\alpha}$ not being cofinal in $(k,<)$ and $(X',<_2)$), so their complements in $X'$ intersect. The set $Y={y\alpha}_{\alpha Y=\{y_\alpha\}_{\alpha<k}$ is what we were looking for.

1