Uncountable ordinals and graphs of functions

Let $$X$$ be the first uncountable ordinal. In other words $$X$$ is an uncountable set equipped with a well-ordering relation "$$\leq$$" such that, for every $$x$$ in $$X$$, the set $$\{y\in X:\ y\leq x\}$$ is countable. Let $$T=\{(x, y)\in X\times X:\ y\leq x \},$$ so that $$T$$ consist of everything below the diagonal in $$X\times X$$.

Let us agree to call a subset $$G\subseteq T$$ a graph, provided $$\big ((x,y_1)\in G\big ) \ \wedge\ \big ((x,y_2)\in G\big ) \ \Rightarrow \ y_1=y_2.$$ Clearly these are precisely the graphs of $$X$$-valued functions $$f$$ defined on subsets of $$X$$, such that $$f(x)\leq x$$, for every $$x$$.

Question: Is it possible to write $$T$$ as the union of a countable family of graphs?

Suppose $$\mathcal{A}=(A_i)_{i\in I}$$ is a collection of countable nonempty sets. Then the set $$[\mathcal{A}]=\{(a,i): a\in A_i\}$$ is the union of countably many graphs of functions.
When phrased this way it's much easier to think about. Simply fix surjections $$b_i:\omega\rightarrow A_i$$, and for $$k\in\omega$$ let $$f_k:I\rightarrow\bigcup\mathcal{A}:i\mapsto b_i(k).$$ Then the union of the graphs of the $$f_k$$s is exactly $$[\mathcal{A}]$$.
(Put another way, $$f_k(i)=b_i(k)$$ - basically, you have $$I$$-many "columns" of size $$\le\omega$$, and you view the whole grid as $$\omega$$-many "rows" of size $$I$$.)
Of course, we use the axiom of choice when we pick the $$b_i$$s, and in the absence of choice can fail (and in particular the special case of your question can fail without choice). But that's a side point.
• Your definition $f_k(i)=b_i(k)$ reminds me that Attila Máté once told a class in combinatorial set theory that the main technique of that subject is interchanging the subscript and the argument. – Andreas Blass May 9 at 22:56