Example of Connected but Not-Path Connected Set $ L:= \{(1,0)\}\;\cup\; \bigcup_{n=1}^\infty L_n$ where $L_n:= \{(x,{x\over n})\in \Bbb R^2\mid \; x\in [0,1]\},\; n\in \Bbb N$.

I've been provided that the set L as an example of Connected but not Path-connecte set.
I would like to prove that the given L is connected but not-path-connected?
Below is my proof for connectedness, please check whether it's correct.
Proof of Connected-ness>>
If we choose two open sets $u,v$ in $\Bbb R^2 $ which are $ u = (1,0)$ and $ v = L \setminus \{(1,0)\} $ then A is connected. 
Proof of Path-Connected >>
How to do this? 
 A: Proof of  connectedness
Let $U,V\subseteq \Bbb R^2$ be arbitrary open sets with $L\subseteq U\cup V$ and $L\cap U\cap V=\emptyset$. We want to show that $L\subseteq U$ or $L\subseteq V$.
One of them, wlog. $U$, contains the point $(1,0)$, hence also an open disk $B_r(1,0)$ of some positive radius $r$ around that point. For $N>\frac1r$, we have $(1,\frac1N)\in U$, hence $U$ intersects $L_N$. As $L_N$ is connected (it is homeomorphic with $[0,1]$), we conclude that $L_n\subseteq U$. In particular $(0,0)\in U$. But the $U$ intersects $L_n$ for all $n$, hence $L_n\subseteq U$ for all $n$, i.e., $\bigcup_{n=1}^\infty L_n\subseteq U$. As also $(1,0)\in U$, we have $L\subseteq U$.
Disproof of path connectedness
Assume $L$ were path-connected. Then there exists a path from $(0,0)$ to $(1,0)$, i.e., a continuos map $\gamma\colon [0,1]\to L$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,0)$.
Let $t_0=\sup\{\,t\in[0,1]\mid\gamma(t)=(0,0)\,\}$. By continuity of $\gamma$, also $\gamma(t_0)=(0,0)$.
Let $t_1=\inf\{\,t\in[t_0,1]\mid\gamma(t)=(1,0)\,\}$. Again by continuity of  $\gamma$, also $\gamma(t_0)=(1,0)$.
Then $p:=\gamma((t_0+t_1)/2)$ is $\ne(0,0)$ and $\ne (1,0)$, hence on one (and only one) $L_n$. But there is no path from $p$ to $(1,0)$ that does not pass through $(0,0)$ (to see this, note that $L\setminus\{(0,0)\}$ is not even connected).
A: Your proof for connectedness does not make sense.  You need to prove that for all pairs $U,V$ of non-empty open subsets of $L$ such that $U\cup V=L$, you have $U\cap V\neq\emptyset$.  Instead, you've only mentioned one choice of $U$ and $V$ - and your set $u$ isn't open anyway.  
Note that the set $L\setminus\{(0,1)\}$ is path connected (just take a path to the origin and then back out again) and hence connected.  Now notice that $(0,1)$ is contained in the closure of $L\setminus\{(0,1)\}$.  You can then apply the following result, which is not too difficult to prove:
Theorem Let $X$ be a connected subset of some space $M$ and suppose that $X\subset Y\subset \text{cl}X$.  Then $Y$ is connected.
To show that your space is not path connected, you will need to show that the point $(0,1)$ is not connected to the rest of the space by a path.  Specifically, let's show that there is no path from $(0,1)$ to $(1,1)$.  Indeed, suppose there were such a path $\gamma\colon[0,1]\to L$.
Let $B$ be the open ball of radius $1/2$ about $(0,0)$.  $L\setminus B$ is not connected (it is easy to find open sets $U,V$ that disconnect it), so any path from $(0,1)$ to $(1,1)$ must pass through $B$.  Let
$$
t=\inf\{s\in[0,1]\colon\gamma(s)\in B\}
$$
We claim that if $u< t$, then $\gamma(u)=(0,1)$.  Indeed, any path from $(0,1)$ to any point in $L\setminus\{(0,1)\}$ must pass through $B$, so if $\gamma(u)\in L\setminus\{(0,1)\}$ then there must be some $v\le u<t$ such that $\gamma(v)\in B$, contradicting the definition of $t$.  Since $\{(0,1)\}$ is closed, it follows that $\gamma(t)=(0,1)$.  
But we also have
$$
t\in\text{cl}\gamma^{-1}(B)\subset\gamma^{-1}(\text{cl}B)
$$
and so $\gamma(t)\in\text{cl}(B)$.  But this is a contradiction, since $(0,1)$ is not in the closure of $B$.  
I think the proof can be tidied up a bit, but that's the idea.
