It is a very nice result ! Let me try an "elementary" answer.

Define $\delta: X\times X \to \mathbb R_+$ by $$\delta(x,y)= \inf_{n\in \mathbb N} d(T^n(x), T^n(y))~.$$ This function is continuous and vanishes exactly on the diagonal by distality. $$\delta(T(x),T(y)) \geq \delta(x,y)~.$$

Set $$F= \bigcap_{n\in \mathbb N} T^n(X)~.$$ $F$ is compact and for all $x\in X$ we have $d(T^n(x), F) \underset{n\to +\infty} \longrightarrow 0$. Since $$\inf_{y\in F} \delta(x,y) \leq \inf_{y\in F} d(T^n(x),y)$$ for all $n$, we conclude that $$\inf_{y\in F} \delta(x,y) = 0~.$$ By compactness of $F$ and distality, we conclude that $x\in F$ for all $x$. Hence $F=X$ and $T$ is surjective.

It is a very nice result ! Let me try an "elementary" answer.

Define $\delta: X\times X \to \mathbb R_+$ by $$\delta(x,y)= \inf_{n\in \mathbb N} d(T^n(x), T^n(y))~.$$ This function is continuous and vanishes exactly on the diagonal by distality.

Edit: This function is NOT continuous in general, as pointed out in the comments.

Set $$F= \bigcap_{n\in \mathbb N} T^n(X)~.$$ $F$ is compact and for all $x\in X$ we have $d(T^n(x), F) \underset{n\to +\infty} \longrightarrow 0$. Since $$\inf_{y\in F} \delta(x,y) \leq \inf_{y\in F} d(T^n(x),y)$$ for all $n$, we conclude that $$\inf_{y\in F} \delta(x,y) = 0~.$$

By compactness of $F$ and distality, we conclude that $x\in F$ for all $x$. Hence $F=X$ and $T$ is surjective.

Edit: Of course this last line falls appart if $\delta$ is not continuous. So the question is: why is the infimum $\inf_{y\in F} \delta(x,y) = 0$ is attained ?