Preference relation $\succsim$ continuous if and only if the upper and lower contour sets are both closed I'm trying to show that a preference relation $\succsim$ is continuous if and only if the upper and lower contour sets are closed.
The direction $\Rightarrow$, i.e. continuity implies the upper and lower contour sets are closed, is trivial, but I'm really struggling to show the other direction. Here is my attempt so far:
Assume that all upper and lower contour sets are closed. Let $x^n$ be a sequence converging to $x$, $y^n$ a sequence converging to $y$ with $x^n \succsim y^n$ for each $n$. Suppose, for a contradiction, that we do not have $x \succsim y$, i.e. $y$ is not contained in $L(x)$, the lower contour set of $x$.
Since $L(x)$ is closed, we have $\bar{L(x)} = L(x)$ and so this means there exists an $\varepsilon>0$ such that $B(y,\varepsilon)\subset X \setminus L(x)$. $y^n \rightarrow y$ so there exists an $N$ such that $y^n\in B(y,\varepsilon)$ for all $n\geq N$. Then I don't know how to proceed.
Note: We have also been given an alternative definition of continuity, which states that if $x^n \rightarrow x$ and $x^n \succsim y$ for each $n$ then $x \succsim y$. It is clear that this is follows from the first definition, but I don't know if the other direction is true. I'm also unsure how this definition would imply closedness of the lower contour sets (the closedness of the upper contour sets is clear).
 A: As you have argued, if the claim is false, then $y^{n} \succ x$ for all but finitely many $n$. 
Analogously, because $U(y)$ is closed, $y \succ x^{n}$ for all but finitely many $n$.
Therefore, for all $n$ larger than some $n_{0}$, we have $y \succ x^{n} \succeq y^{n} \succ x$.
In particular, $y \succ y^{n_{0}} \succ x$.
See if you can finish the proof from here by using the fact that the upper and lower contour sets of $y^{n_{0}}$ are closed. (Just to be sure, there is no special reason for using $y^{n_{0}}$ instead of $x^{n_{0}}$ to complete the proof. The argument given below only requires that there be some point $z$ in $X$ that satisfies $y \succ z \succ x$.) 
Using closedness of the upper and lower contour sets again, we infer that  there is an open neighborhood $O_{y}$ of $y$ contained in $X \setminus L(y^{n_{0}})$, and an open neighborhood $O_{x}$ of $x$ contained in $X \setminus U(y^{n_{0}})$.
Since the two sequences are assumed to converge to $x$ and $y$ respectively, there  exists a number $n_{1}$ such that for all $n$ greater than $n_{1}$,
\begin{align*}
y^{n} \in O_{y} \subseteq X \setminus L(y^{n_{0}}) \quad \text{and}\quad
x^{n} \in O_{x} \subseteq X \setminus U(y^{n_{0}}).
\end{align*}
But now we have reached a contradiction, since $y^{n} \succ y^{n_{0}} \succ x^{n}$ is impossible if $x^{n} \succeq y^{n}$ for all $n$.

The second definition does not imply the first. 
For a counterexample, let the relation to be such that it is represented by a function $f$ which is upper-semicontinuous, but not lower-semicontinuous.
For instance, let $X = [0, 1]$ and let $f\colon[0, 1]\to\mathbb{R}$ be the function 
$$
f(x) = \begin{cases}
-x \quad \text{, if $x \in [0, 1/2)$},\\
1 + x \quad \text{, else}.
\end{cases}
$$
Then $x^{n} \succeq y$ and $x^{n} \to y$ imply $x \succeq y$ since $f$ is upper-semicontinuous.
On the other hand, the lower contour sets are not closed. To see this, note that $0 \succeq 1/2 - 1 / n$ for every $n$ since $f(0) \geq f(1/2 - 1/n)$. However, $1/2 \succ 0$ since $f(1/2) > f(0)$.
