Here is (a bit lengthy and technical) proof of the Lindelof property of the function space $C_k(\omega_1)$. At first some notations.

For any function $f\in C_k(\omega_1)$ and a countable ordinal $\alpha$ let $\|f\|_\alpha=\sup_{x\in[0,\alpha]}|f(x)|$. Let also $\|f\|=\sup_{x\in\omega_1}|f(x)|$.
For every $f\in C_k(\omega_1)$, $\alpha\in\omega_1$ and $\varepsilon>0$ consider the open neighborhood $$B_\alpha[f;\varepsilon):=\{g\in C_k(\omega_1):\|g-f\|_\alpha<\varepsilon\}$$ of $f$ in the function space $C_k(\omega_1)$.

Given an open cover ${\mathcal U}$ of $C_k(\omega_1)$, for every $f\in C_k(\omega_1)$ let $$
\begin{aligned}
\varepsilon_f&:=\sup\{\varepsilon\in (0,1]:\exists \alpha\in\omega_1\;\exists U\in{\mathcal U}\;\;B_\alpha[f;4\varepsilon)\subset U\},\mbox{ and }\\
\alpha_f&:=\min\{\alpha\in\omega_1:\exists U\in{\mathcal U}\;\;B_\alpha[f,3\varepsilon_f)\subset U\}.
\end{aligned}
$$
Choose also a set $U_f\in{\mathcal U}$ such that $B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.

**Claim.** *For any functions $f,g\in C_k(\omega_1)$ we get $4\varepsilon_f\ge 4\varepsilon_g-\|f-g\|$.*

*Proof.* Assuming that $4\varepsilon_f<4\varepsilon_g-\|f-g\|$, we can choose $\delta>0$ such that $\|f-g\|+4\varepsilon_f+\delta\le 4\varepsilon_g-\delta$ and find an ordinal $\alpha\in\omega_1$ such that $B_\alpha[g;4\varepsilon_g-\delta)\subset U$ for some $U\in{\mathcal U}$. Then $$B_\alpha[f;4\varepsilon_f+\delta)\subset B_\alpha[g;\|f-g\|+4\varepsilon_f+\delta)\subset B_\alpha[g;4\varepsilon_g-\delta)\subset U\in{\mathcal U},$$ which contradicts the definition of the number $\varepsilon_f$. $\square$

For every ordinal $\alpha\in\omega_1$, identify the Banach space $C_k[0,\alpha]$ with the subspace $\{f\in C_k(\omega_1):f|[\alpha,\omega_1)\equiv const\}$ of $C_k(\omega_1)$, consisting of functions, which are constant on the interval $[\alpha,\omega_1)$.
For two ordinals $\alpha,\beta$ by $\alpha{\vee}\beta$ we denote their maximum $\max\{\alpha,\beta\}$.

We shall construct inductively a non-decreasing sequence of countable ordinals $(\alpha_n)_{n\in\omega}$ and a sequence $(F_n)_{n\in\omega}$ of countable subsets $F_n\subset C_k[0,\alpha_n]$ such that for every $n\in\omega$ the following conditions are satisfied:

$(1_n)$ $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$;

$(2_n)$ $\alpha_{n+1}=\sup\limits_{f\in F_n}(\alpha_n{\vee}\alpha_f)$.

We start the inductive construction letting $\alpha_0=0$. Assume that for some $n\in\omega$ an ordinal $\alpha_n$ has been constructed. For the open cover $\{B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f):f\in C_k[0,\alpha_n]\}$ of the separable Banach space $C_k[0,\alpha_n]$, there exists a countable subset $F_n\subset C_k[0,\alpha_n]$ such that $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n{\vee}\alpha_f}[f;\varepsilon_f)$. Letting $\alpha_{n+1}:=\sup_{f\in F_n}(\alpha_n{\vee}\alpha_f)$ we complete the inductive step.

After completing the inductive construction, consider the countable ordinal $\alpha_\omega=\sup_{n\in\omega}\alpha_n$ and the countable set $F:=\bigcup_{n\in\omega}F_n\subset C_k[0,\alpha_\omega]\subset C_k(\omega_1)$. We claim that $\{U_f:f\in F\}\subset {\mathcal U}$ is a countable subcover of $C_k(\omega_1)$. Given any function $g\in C_k(\omega_1)$, for every ordinal $n\le\omega$ consider the (unique) function $g_n\in C_k[0,\alpha_n]$ such that $g_n|[0,\alpha_n]=g|[0,\alpha_n]$.
By the continuity of the function $g$ at $\alpha_\omega$, there exists a number $n\in\omega$ such that $|g(x)-g(\alpha_\omega)|<\varepsilon_{g_\omega}$ for all $x\in[\alpha_n,\alpha_\omega]$. This implies that $\|g_n-g_\omega\|<\varepsilon_{g_\omega}$ and $4\varepsilon_{g_n}\ge 4\varepsilon_{g_\omega}-\|g_n-g_\omega\|>3\varepsilon_{g_\omega}$ according to the Claim.

By the inductive condition $(1_n)$, for the function $g_n\in C[0,\alpha_n]$ there exists $f\in F_n$ such that $g_n\in B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$ and hence $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$. By the inductive condition $(2_n)$, $\alpha_f\le\alpha_{n+1}\le\alpha_\omega$. Since the functions $f$ and $g_n$ are constant on the interval $[\alpha_n,\omega_1)$, the inequality $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$ implies $\|f-g_n\|=\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}$. Then Claim yields that $4\varepsilon_{f}>4\varepsilon_{g_n}-\|f-g_n\|>3\varepsilon_{g_n}$ and hence $\varepsilon_{g_\omega}<\frac43\varepsilon_{g_n}<\frac43\frac43\varepsilon_{f}$.

We claim that $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$. Indeed, for every $x\in [0,\alpha_f]$ we get $$|g(x)-f(x)|\le |g(x)-g_n(x)|+|g_n(x)-f(x)|<\varepsilon_{g_\omega}+\varepsilon_f<\tfrac{16}9\varepsilon_{f}+\varepsilon_f<3\varepsilon_f,$$
and hence $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.