FALSE: every countable language is decidable

The set $\{1{\}}^{\ast }$ is countable by length, so write its elements as $w_n:=1^n$ for $n\ge 0$. Filter the effective list from [L2] by keeping only the coded machines whose input-alphabet parameter satisfies $s\ge 1$; by [L4], this gives an effective enumeration $N_0,N_1,N_2,\dots$ of all coded deterministic one-tape machines that can read every unary word. Let $D:=\{w_n:N_n\text{ does not accept }{w}_n\}$. By [L1], the language $D$ is countable because it is a subset of the countable set $\{1{\}}^{\ast }$.

Suppose a decider decides $D$. By [F1], an equivalent coded decider occurs as some $N_k$ in the filtered enumeration. If $w_k\in D$, then step 1.1 says that $N_k$ does not accept $w_k$, contradicting the requirement that a decider for $D$ accept members. If $w_k\mathrm{\notin }D$, then step 1.1 says that $N_k$ accepts $w_k$, contradicting the requirement that a decider reject nonmembers.

Thus no decider decides the countable language $D$, so the statement is false.