For a sequence $X_n$ of real numbers with $\limsup\limits_n X_n = \inf\limits_n\sup\limits_{k\geqslant n} X_k$, $\liminf\limits_n X_n = \sup\limits_n \inf\limits_{k\geqslant n} X_k$,
(a) How to prove that $X_n$ tends to $x$ is equivalent to saying that $\limsup_n X_n = \liminf_n X_n = x$
(b) If $X_n \geqslant Y_n$ for all $n \geqslant 1$ and if one of them is convergent, how to prove that $\liminf_n X_n \geqslant \limsup_n Y_n$. In general, $\liminf_n X_n \geqslant \liminf_n X_n;\; \limsup_n X_n \geqslant \limsup_n Y_n$
b) is simpler than it looks. If $x_n \ge y_n$ for all $n$ then $\liminf x_n \ge \liminf y_n$ and $\limsup x_n \ge \limsup y_n$. If (for instance) $y_n$ is convergent, then $\liminf y_n = \limsup y_n$ so you get the inequality you need.
For a) suppose the liminf and limsup are equal to $x$. Let $y_n = \sup_{k \ge n} x_k$ and $z_n = \inf_{k \ge n} x_k$. Then $y_n \searrow x$ and $z_n \nearrow x$. Let $\epsilon > 0$ be given. There exist indices $N_y$ and $N_z$ with the property that $n \ge N_y \implies x \le y_n < x + \epsilon$ and $n \ge N_z$ implies $x-\epsilon < z_n \le x$. If $N = \max\{N_y,N_z\}$ then $$n \ge N \implies x-\epsilon < y_n \le x_n \le z_n < x + \epsilon.$$ It follows that $x_n \to x$. The converse is proven in a similar way.