Diagonalizing against $\omega_1$-sequences of functions mod finite The following statement is a direct consequence of the Continuum Hypothesis:

There exists a sequence $\langle f_\alpha:\omega_1\rightarrow\omega_1 ~ \vert ~ \alpha<\omega_1\rangle$ of functions such that there is no function $f:\omega_1\rightarrow\omega_1$ with the property that the sets $\{\xi<\omega_1 ~ \vert ~ f(\xi)=f_\alpha(\xi)\}$ are finite for all $\alpha<\omega_1$.

Moreover, since a failure of this statement can be used to obtain an $\omega_2$-sequence  of subsets of $\omega_1$ with pairwise finite intersection, results of Baumgartner in
Baumgartner, James E., Almost-disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 9, 401-439 (1976). ZBL0339.04003.
show that the statement is not equivalent to CH.
Question: Can the above statement consistently fail?
 A: The arguments in Section 6 of my paper "The nonstationary ideal in the $\mathbb{P}_{\mathrm{max}}$ extension" show that there is a proper forcing adding a function from $\omega_{1}$ to $\omega_{1}$ which agrees with each such ground model function in only finitely many places. I say there that Todorcevic had done something similar in his "A note on the Proper Forcing Axiom". It follows that under PFA, and in the $\mathbb{P}_{\mathrm{max}}$ extension, the statement above fails. You don't need any large cardinals, however.
A: This isn't an answer, as you're working in ZFC. But it seems worth noting.
Assume ZFC + AD$^{L(\mathbb{R})}$. Then
$L(\mathbb{R})$ satisfies ZF + AD + DC + "the statement is false".
Proof: Work in $L(\mathbb{R})$. Then every real has a sharp, and every subset of $\omega_1$ is
constructible from a real. Suppose  $\vec{f}=\left<f_\alpha\right>_{\alpha<\omega_1}$
witnesses Statement. Let $x\in\mathbb{R}$
be such that $\vec{f}\in L[x]$. Define $g:\omega_1\to\omega_1$ by
$g(\alpha)=$ the $(\alpha+1)$th $x$-indiscernible. So $g$ is
injective, strictly increasing, and its range consists of
$x$-indiscernibles. Suppose $\beta<\omega_1$ and $A\subseteq\omega_1$
is infinite and $f_\beta\upharpoonright A=g\upharpoonright A$. Fix a
finite set $s$ of $x$-indiscernibles such that $f_\beta$ is definable
in $L[x]$ from $(s,x)$. Since $g``A$ is infinite, we can choose
$\iota\in(g``A)\backslash s$.
Let $\xi=g^{-1}(\iota)$. Since $f_\beta(\xi)=\iota$, $\iota$ is
definable over $L[x]$ from $(s,\xi,x)$. But $\xi<\iota$, so we can
choose a finite set $t$ of $x$-indiscernibles such that $\xi$ is
definable from $(t,x)$, with $\iota\notin t$. So $\iota$ is definable
over $L[x]$ from $(s\cup t,x)$, where $s\cup t$ are $x$-indiscernibles
and $\iota\notin s\cup t$, and this contradicts that $\iota$ is an
$x$-indiscernible.
Remark: The $\mathbb{P}_{\mathrm{max}}$ extension of $L(\mathbb{R})$  inherits some related properties, and it seems tempting to try to do a version of the preceding argument there, but I haven't seen how to do that.
