Topologist's sine curve is not path connected

This example is to show that a connected topological space need not be path-connected.

The definition I have for X being path-connected is that for any 2 points of X, there is a path connecting them. So if the continuous map p:[0,1]-->X is the path connecting x and y in X, we have p(0)=x; p(1)=y.

S={ (t,sin(1/t)): 0 <t <= 1 }

A={ (0,t): -1 <= t <= 1 }

let T = S U A

with the topology induced from R^2.

I show T is not path-connected. Assume to the contrary that there exists a path p:[0,1]-->T with p(0)=(1/pi, 0) and p(1)=(0,0).

(Here I chose the points x=(1/pi, sin(1/(1/pi)) )= (1/pi, 0) and y = (0,0) in T )

Let k = Inf {t in [0,1] : p(t) in A}.

Then p([0,k]) contains at most one point of A. I want to show A is contained in the Closure of p([0,k]) . So p([0,k]) is not closed, and therefore not compact. But p is continuous and [0,k] is compact, so p([0,k]) must be compact,which is a contradiction.

How do I show Closure of p([0,k]) contains all of A?