Dove tailing questions about accepting language
Showing some machine $R$ that recognizes $L$ is enough to prove that $L$ is recognizable, however, you have yet shown $R$ recognizes $L$.

Formally, you must show:


$M \in L \Rightarrow  R $ accepts $M$
$M \notin L \Rightarrow  R $ does not accept $M$, but not necessarily halts.




Let $S_i$ be all strings in $\Sigma ^*$ of length exactly $i$. It is clear, for any $i$ that is finite, $S_i$ is also finite.

Let $S_i^*$ be a subset of $S_i$, where every $s \in S_i ^*$ contains exactly five $1$'s.

$R$ is a machine that given input $M$:


for $i=5...\infty$
for $s$ in $S_5^* \cup S_6^* \cup \dots \cup S_i^*$

simulate $s$ on $M$ for $i$ steps. If $M$ halts in $i$ steps or less, finish and accept. Otherwise continue.


Suppose $M \in L $ (and we will show $R$ accepts $M$):
 There exists $w^*$ of length $k$ that contains five $1's$ and $M$ halts on $w^*$ in $r$ steps.

Since $S_i^*$ is finite, $S_5^* \cup S_6^* \cup \dots \cup S_k^*$ is also finite (finite union of finite subsets), so $R$ simulates $w^*$ on $M$. Now if $r \leq k$ then $M$ halts and $R$ accepts $M$.

If $r &gt; k$ then surely $w^* \in S_5^* \cup S_6^* \cup \dots \cup S_r^*$, hence at $i=r$ $M$ will halt on $w^*$ and $R$ will accept.

The second step is very simple; if $M \notin L$ then no such $w$ with five $1's$ exists such that $M$ halts on it, and $R$ will run indefinitely.