Let $X\subset \mathbb{R}^2$ be a compact set and $(x_n,y_n)_{n\in \mathbb{N}}\subset X\;$ be a sequence. Assume that $x_n$ converges to $x^*$. By Bolzano-Weierstrass it is known that there is a subsequence $(x_{n_k},x_{n_k})_{k\in\mathbb{N}}\;$ which converges to $(x^*,y^*:=\liminf_{k\rightarrow \infty}\;y_{n_k})\in X$ as it is clearly a limit point of $(x_n,y_n)_{n\in \mathbb{N}}\;$.
I would like to show that if $x_n\rightarrow x^*$ then$$\liminf_{n\rightarrow \infty}y_n \geq y^* .$$
By definition we have $$\liminf_{k\rightarrow \infty}y_{n_k}= y^*,$$ but in general we have $$y^*=\liminf_{k\rightarrow \infty}y_{n_k}\geq \liminf_{n\rightarrow\infty}y_n.$$What am I missing, or is it even possible to show what I want to show? To be clear I only need $\liminf_{n\rightarrow \infty}\; y_n$ to be equal or greater than some arbritary $y^*$ where $(x^*,y^*)\in X$. Thank you.
I don't quite understand your question, since if the subsequence $(y_{n_k})_{k\in\mathbb{N}}$ is chosen to be convergent, then $$ y^\ast=\liminf_{k\to\infty}y_{n_k}=\lim_{k\to\infty}y_{n_k}, $$ so no need to consider limes inferior, but you cannot a priori define $y^\ast=\liminf_{k\to\infty}y_{n_k}$, since you have not specified the subsequence $(y_{n_k})_{k\in\mathbb{N}}$. Maybe you meant to write $y^\ast=\liminf_{n\to\infty}y_n$, but then the rest of the question doesn't really make sense.
I will try to answer what I think you are asking. Feel free to comment if I have misunderstood! Let's ignore $(x_n)_{n\in\mathbb{N}}$ and $x^\ast$ and just take some arbitrary subsequence $(y_{n_k})_{k\in\mathbb{N}}$ of $(y_n)_{n\in\mathbb{N}}$. Here, $(y_n)_{n\in\mathbb{N}}$ may be any sequence of real numbers. Then, for each $N\in\mathbb{N}$, we clearly have $$ \{ y_{n_k}\mid k\in\mathbb{N},k\geq N\}\subseteq\{ y_n\mid n\in\mathbb{N},n\geq N\}, $$ and thus $$ \inf\{ y_{n_k}\mid k\in\mathbb{N},k\geq N\}\geq\inf\{ y_n\mid n\in\mathbb{N},n\geq N\}. $$ Letting $N\to\infty$, we get $$ \liminf_{k\to\infty}y_{n_k}\geq\liminf_{n\to\infty}y_n. $$