I am trying to prove the following:
Show that local connectedness is not a continuous invariant by describing a continuous function $f:[0,1] \rightarrow Y$ from $[0,1)$ unto the space Y shown below. 
Here is what I have done so far, but I am not sure whether this is rigorous enough:
Let the dot be point $a$ and let the curves coming from the stop at point $b$. $[0,1)$ is a locally connected space. Let $f:[0,1] \rightarrow Y$ be such that the path $f$ begins at $a$ and terminates at $b$. But then $a$ is not locally connected since any open set containing $a$ that also contains $b$ (that also does not contain the entire space of Y) does not contain any connected open sets. This is because that since $a$ and $b$ are so close that any open set containing $a$ must also contain $b$. Thus, local connectedness is not a continuous invariant.
Can anyone help me with making this proof more rigorous?
Thanks!
This question comes from Principles of Topology by Fred Croom, chapter 5.6, #12.
There is no point $b$: the wavy part of the curve does not have a left endpoint. You need to define $f$ piecewise, in five pieces. The first piece could be $f(x)=\langle -5x,0\rangle$ for $0\le x\le\frac15$: that maps $\left[0,\frac15\right]$ onto the segment of the $x$-axis from $\langle 0,0\rangle$ to $\langle -1,0\rangle$. The second piece could then map $\left[\frac15,\frac25\right]$ onto the vertical segment from $\langle -1,0\rangle$ down to $\langle -1,-2\rangle$:
$$f(x)=\left\langle-1,-10\left(x-\frac15\right)\right\rangle=\langle-1,2-10x\rangle\,.$$
The third piece could then map $\left[\frac25,\frac35\right]$ onto the horizontal segment from $\langle-1,-2\rangle$ to $\langle 1,-2\rangle$, the fourth then mapping $\left[\frac35,\frac45\right]$ onto the vertical segment from $\langle 1,-2\rangle$ up to $\langle 1,\sin1\rangle$, while the fifth could finish the job by mapping $\left[\frac45,1\right)$ onto topologist’s sine curve. For that last piece the map could be
$$\begin{align*} f(x)&=\left\langle 1-5\left(x-\frac45\right),\sin\frac1{1-5\left(x-\frac45\right)}\right\rangle\\ &=\left\langle5-5x,\sin\frac1{5-5x}\right\rangle\,. \end{align*}$$
I’ll leave it to you to work out the details for the third and fourth pieces and to convince yourself that the resulting function $f$ really does wrap the interval $[0,1)$ around the space $Y$, starting at $\langle 0,0\rangle$ (your $a$) and going counterclockwise around $Y$.