The graph of a function $f : X → Y$ refers to the set $[f] :=$ {$(x, f(x)) : x ∈ X$} $⊆ X × Y$.
Show that a function $f : \mathbb R → \mathbb R$ is continuous if and only if its graph $[f]$ is path-connected.
My attempt:
Suppose {$x_n$} is a sequence in $\mathbb R$ converging to some $x$. We will show that there is a subsequence so that $f(x_{n_k})→f(x)$. This is sufficient; for if $f$ were not continuous at $x$, there would exist an $ϵ$ such that for any $n$ we could find an $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)−f(x)|>ϵ$. Then we would have $x_n→x$, but no subsequence of {$f(x_n)$} could converge to $x$, contradicting our claim.
Since the graph of $f$ is path connected, there is a continuous path $c(t)=(u(t),v(t))$ with $v(t)=f(u(t))$ for each $t$, and $c(0)=(a,f(a))$, $γ(1)=(b,f(b))$. By the intermediate value theorem, for each $n$ there is a $t_n∈[0,1]$ with $u(t_n)=x_n$. Since $[0,1]$ is compact, we can find a subsequence {$t_{n_k}$} converging to some $t$. Then by continuity of $u$, since $u(t_{n_k})=x_{n_k}→x$, we have $u(t)=x$. Now $v$ is also continuous, so $f(x_{n_k})=f(u(t_{n_k}))=v(t_{n_k})→v(t)=f(u(t))=f(x)$. We have thus constructed the desired subsequence.
How to show the other conclusion
For the converse, the map $ \gamma: x \mapsto (x,f(x))$ is continuous as both of its projections are continuous if $f$ is supposed to be continuous.
If $P_1=(x_1, f(x_1)), P_2=(x_2, f(x_2)))$ are in the graph $\Gamma(f)$ of $f$
$$g(t)=\gamma((1-t)x_1+tx_2)$$ is a continuous path in $\Gamma(f)$ from $P_1$ to $P_2$. Proving that $\Gamma(f)$ is path connected.