Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to both sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly.
PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in {q/2, q}$\lbrace q/2, q\rbrace$,$\zeta^j$is non-real, so the corresponding parametrization does not give any real points. 2 added 300 characters in body Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let$C$be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of$x$) of$C$near$O$. By assumption one of the branches (over$\mathbb{C}$), call it$C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for$a_i \in \mathbb{R}$. Let$q$be the least common multiple of the denominators of$r_i$'s. If$q$is odd, then the branch expands to both sides of the origin and therefore$C$C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly.
PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in {q/2, q}$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.
Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to both sides of the origin and therefore $C$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly.