Here $X$ is defined as the set $\{(x,\frac{1}{x}\sin(1/x)) : x>0\} \cup\{ \text{y axis}\}$
Here is a sketch: Suppose there is such a $\gamma$ mapping $[0,1]$ to $X$, with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$. Then by IVT on the first coordinate, we have all points in $[0,1]$ being hit by the first coordinate. So construct sequences $x_n$, $y_n$ $\rightarrow 0 $ with $\sin(x_n)=0$, $\sin(y_n)=1$. We must have sequences $p_n, q_n \rightarrow 0$ with $\gamma(p_n)=(x_n,0)$ and $\gamma(q_n)=(y_n,1/y_n)$ ie no unique limit, contradicting continuity of $\gamma$
Edit: I'm pretty sure now it is not, and the flaw is in existence of sequences p and q.
Your definition of the topologist's sine curve is somewhat unusual, but it is certainly a variant of it. The "standard" definition is $$T = \{(x,\sin(1/x)) : x>0\} \cup \left( \{ 0 \} \times [-1,1] \right)$$ which is compact.
The idea of your proof is correct, but it is not really clear how you define and use the sequences $(x_n), (y_n)$.
You assume that there is a map $\gamma : [0,1] \to X$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$ and invoke the IVT to show that $\gamma_1 = p_1 \circ \gamma : [0,1] \to [0,1]$ is surjective (where $p_1 : X \to [0,1]$ is the projection onto the first coordinate).
I suggest to proceed as follows. Define $x_n = (\pi/2 + 2n \pi)^{-1}$. Then $x_n \to 0$, $\sin (1/x_n)= 1$ and $\frac{1}{x_n} \sin(1/x_n) \to \infty$. Choose $t_n \in[0,1]$ such that $\gamma_1(t_n) = x_n$. The sequence $(t_n)$ must have a convergent subsequence $(t_{n_k})$ with limit $\tau \in [0,1]$. Since $\gamma$ is continuous, the sequence $(\gamma(t_{n_k}))$ converges to $\gamma(\tau)$. Because $\gamma_1(t_{n_k}) = x_{n_k} > 0$, we conclude that $\gamma(t_{n_k}) = (x_{n_k},\frac{1}{x_{n_k}} \sin(1/x_{n_k}))$. But this is a contradiction because $\frac{1}{x_{n_k}} \sin(1/x_{n_k}) \to \infty$.