How to show that graph of a continuous function on an interval is pathwise-connected. 
Can anybody help me in solving this problem.
Thanks.
My attempt:
Define $g:I \rightarrow \mathbb R^2$ as $g(t)=(t,f(t))$, then $g$ is continuous as all its components are continuous. Since $I$ is compact and connected then the image of $g$ must also be compact and connected and the image of $g$ in $\mathbb R^2$ is exactly the graph of $f$.
But I got connectedness, how to move to pathwise-connectedness.
Is this the right approach?
Definition I am using 
Pathwise connectedness: A set is said to be path connected if for any two points in the set there is a path in the set joining those two points.
Path: Continuous function from $[0,1]\rightarrow A$  where $f(0)=x$and $f(1)=y$ is a path from x to y. 
 A: It is sufficient to show that we can define a continuous path connecting any two points in $G$. Let two distinct points $p,q\in G$ be given. Here, $p=(p_x, f(p_x))$ and $q=(q_x,f(q_x))$ for appropriately chosen points $p_x,q_x\in I$. Assume without loss of generality that $p_x<q_x,$ then consider the closed interval $[p_x,q_x]$.
We know that there is a continuous function $\phi: I\to [p_x,q_x].$ Now define a function $F$ using the restriction of $f$ to $[p_x,q_x]\subseteq I$.
$$ F:[p_x,q_x]\mapsto G$$ 
$$F(x)=(x,f(x)).$$
$F$ maps $[p_x,q_x]$ continuously onto a path $[p,q]\in G.$ This follows from continuity of $f$. Moreover, $F(p_x)=(p_x,f(p_x))$ and $F(q_x)=(q_x,f(q_x)).$ Finally, define the composition of functions
$$ F\circ\phi: I\to G.$$
$F$ and $\phi$ are both continuous so their composition is continuous. Their composition maps $I$ to a continuous path in $G$, such that the beginning and end points are any two points in $G$ that you wish. Thus, every two points can be connected by a path and $G$ is path connected.
$G$ is convex if and only if the graph of the function is a straight line.
