Show that $V = \{\langle x, y \rangle \ | \ x \geq 0 \ \text{and} \ y \in \mathbb{R}\}$ is closed. 
Show that $V = \{\langle x, y \rangle \ | \ x \geq 0 \ \text{and} \ y \in \mathbb{R}\}$ is closed.

My Attempted Proof: 
We claim all of the boundary points of $V$ lie on the line $x = 0$. To prove this claim, pick a point $\alpha = \langle 0, y \rangle$ where $y \in \mathbb{R}$, along the line $x = 0$. Let $U$ be a neighbourhood of $\alpha$. Since $U$ is an open set and $\mathbb{R}^2$ is a metric space, there exists an open ball $B = B_{(\mathbb{R}^2, d)}(\alpha, \epsilon) \subseteq U$, where $d$ is the standard metric on $\mathbb{R}^2$ for some $\epsilon > 0$. 
It is clear that $B$ contains a point in $V$ and a point outside of $V$, for example $\langle \frac{\epsilon}{2}, y \rangle$ and $\langle \frac{-\epsilon}{2}, y \rangle$ respectively. Since $U$ was an arbitrary neighbourhood of $\alpha$, every neighbourhood of $\alpha$ intersects $V$ in at least one point in $V$ and one point outside of $V$. Hence $\alpha$ is a boundary point of $V$, and also $\alpha \in V$.
We now show that those points along the line $x = 0$ are the only boundary points of $V$. Pick any point $\beta = \langle \delta, y \rangle$, for some fixed $y >0$ and some fixed $\delta > 0$.  The neighbourhood $W = B_{(\mathbb{R}^2, d)}(\beta, \frac{\delta}{4})$ contains only points in $V$. Hence $\beta$ cannot be a boundary point of $V$.
Now since $\text{Bd}(V) \subseteq V$, we have that $V$ is a closed set in $(\mathbb{R}^2, d)$. $\square$

First of all, is this proof satisfactory and rigorous enough? Secondly, I've only proved this using the fact that if $X$ is a topological space and $A \subseteq X$ is a subset, then $\bar{A} = \text{Int}(A) \cup \text{Bd}(A)$. Is there any easier way to prove it? 
I tried proving it by claiming that $V$ contained all of it's limit points, which I tried to do by letting $(x_n)_{n \in \mathbb{N}}$ be a convergent sequence in $(\mathbb{R}^2, d)$, where $x_n \in V$ for each $n \in \mathbb{N}$, and supposing that $p \in \mathbb{R}^2$ was a limit point and showing that $p \in V$, but I ran into some trouble. How could I complete a proof using this method?
 A: Nicely done! I don't see any issues with your proof.
As far as proving that all limit points of $V$ are elements of $V,$ I'd suggest using topological limit points instead of sequential limit points. In other words, take any $\vec v\in\Bbb R^2$ such that for all $\epsilon>0,$ $B_{(\Bbb R^2,d)}\bigl(\vec v,\epsilon\bigr)$ contains a point of $V$ distinct from $\vec v.$ Then, show that $\vec v$ must be an element of $V$ (proof by contradiction may be your best bet, here).
Another, even easier approach is this: show that $\Bbb R^2\setminus V$ is open. This is probably the "best" approach, as it can be proved simply, directly, and constructively.
A: The easiest way to prove this is probably by using the fact that, under a continuous map, the inverse image of an open set is open, and the inverse image of a closed set is closed. The projection function $\pi:(x,y)\mapsto x$ is continuous, and your set $V$ is the preimage of $[0,\infty)$, which is closed in $\Bbb R$.
Alternatively, suppose $(x,y)\not\in V$, so $x=-a$ for some $a>0$. Then the ball around $(x,y)$ with radius $\frac{a}2$ is contained in the complement of $V$, thus showing that the complement is open, so $V$ is closed.
This second approach is also how you could make the sequential argument work. If a sequence $(x_n,y_n)$ is in $V$, and if it approaches a limit not in $V$, then it approaches some $(x,y)$ where $x=-a$. Thus, the $x$-values in your sequence are a sequence of non-negative real numbers whose limit is a negative real number - impossible.
Your proof using the boundary seems to work as well. Again, this hinges on the fact that points not on the line $x=0$ can be bounded away from that line with an open ball.
A: This is essentially Tony Jacobs alternative, rephrased  a bit.
Correct me if wrong:
$V^c =( \mathbb{R}×\mathbb{R}$\ $ V$ ) $=$
{$(x,y)  | x<0, y \in \mathbb{R} $}.
Show that $V^c$ is open:
Let $(x,y) \in V^c.$
Let $r:= |x| /2.$
Then 
$B_r(x,y) = $
{$(x',y')$| $(x'-x)^2 + (y'-y)^2< r^2$}.
$B_r(x,y)$  $\subset$  $V^c.$
Hence $V^c$ is open, implies $V$ closed.
