A relevant idea by @sure was relegated (by her/him-self!) to comments. Let me give it justice here. Actually, I'll provide a counter-example:

TTHEOREM   There exists a non-continuus map $\ f : X\ \rightarrow Y\ $ of a metric 1-dimensional compact connected space $\ X\ $ into a metric 1-dimensional separable complete space $\ Y\ $ such that the graph $\ G(f)\subseteq X\times Y\ $ is a closed connected subset.

PROOF   Let $\ X:=S^1\ $ be the unit circle in the complex plane, with center $\ 0.\ $ Let $\ Y:=\mathbb R.\ $ Define

$$\forall_{t\,:\, 0\le t< 2\cdot\pi}\ \ f\left(e^{\imath\cdot t}\right)\ :=\ \tan\left(\frac t4\right)$$

That's it, END of Proof.

A relevant idea by @sure was relegated (by her/him-self!) to comments. Let me give it justice here. Actually, I'll provide a counter-example:

THEOREM   There exists a non-continuous map $\ f : X\ \rightarrow Y\ $ of a metric 1-dimensional compact connected space $\ X\ $ into a metric 1-dimensional separable complete space $\ Y\ $ such that the graph $\ G(f)\subseteq X\times Y\ $ is a closed connected subset.

PROOF   Let $\ X:=S^1\ $ be the unit circle in the complex plane, with center $\ 0.\ $ Let $\ Y:=\mathbb R.\ $ Define

$$\forall_{t\,:\, 0\le t< 2\cdot\pi}\ \ f\left(e^{\imath\cdot t}\right)\ :=\ \tan\left(\frac t4\right)$$

That's it, END of Proof.

A relevant idea by @sure was relegated (by her/him-self!) to comments. Let me give it justice here. Actually, I'll provide a counter-example:

TTHEOREM   There exists a non-continuus map $\ f : X\ \rightarrow Y\ $ of a metric 1-dimensional compact connected space $\ X\ $ into a metric 1-dimensional separable complete space $\ Y\ $ such that the graph $\ G(f)\subseteq X\times Y\ $ is a closed connected subset.

PROOF   Let $\ X:=S^1\ $ be the unit circle in the complex plane, with center $\ 0.\ $ Let $\ Y:=\mathbb R.\ $ Define

$$\forall_{t\,:\, 0\le t\le 2\cdot\pi}\ \ f\left(exp\left(\imath\cdot t\right)\right)\ :=\ \tan(\frac t4)$$

That's it, END of Proof.

A relevant idea by @sure was relegated (by her/him-self!) to comments. Let me give it justice here. Actually, I'll provide a counter-example:

TTHEOREM   There exists a non-continuus map $\ f : X\ \rightarrow Y\ $ of a metric 1-dimensional compact connected space $\ X\ $ into a metric 1-dimensional separable complete space $\ Y\ $ such that the graph $\ G(f)\subseteq X\times Y\ $ is a closed connected subset.

PROOF   Let $\ X:=S^1\ $ be the unit circle in the complex plane, with center $\ 0.\ $ Let $\ Y:=\mathbb R.\ $ Define

$$\forall_{t\,:\, 0\le t< 2\cdot\pi}\ \ f\left(e^{\imath\cdot t}\right)\ :=\ \tan\left(\frac t4\right)$$

That's it, END of Proof.