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Edit: I improved the constant to $c = \frac{2}{3}$.

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$} $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

Proof. Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$. For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.

Edit: I improved the constant to $c = \frac{2}{3}$. (Later edit: But the optimal constant turns out to be $c = \frac{1}{2}$, see Yuval Peres' answer.)

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$} $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

Proof. Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$. For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.

Edit: I improved the constant to $c = \frac{2}{3}$.

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \qquad (*) $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

Proof. Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$. For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in $(*)$ is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.

Edit: I improved the constant to $c = \frac{2}{3}$.

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$} $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

Proof. Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$. For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{7}{9} M, \qquad (*) $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{7}{9}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances strictly larger than $\frac{7}{9}$.

Proof. Assume that all three distances are strictly larger than $\frac{7}{9}$ and set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$.

For $k \in \{1,2,3\}$, let $a_k$ be the measure of $\{g_k > \frac{1}{3}\}$ and $b_k = 1-a$ the measure of $\{g_k \le \frac{1}{3}\}$. Then $$ \frac{7}{9} < \int g_k \le a_k + \frac{1}{3}b_k = \frac{1}{3} + \frac{2}{3}a_k, $$ so $a_k > \frac{2}{3}$. Hence, the set in $[0,1]$ where all three functions $g_1,g_2,g_3$ are strictly larger than $\frac{1}{3}$ has non-zero measure.

But for $x$ from this set, the three numbers $f_1(x), f_2(x), f_3(x) \in [0,1]$ have mutual distance strictly larger than $\frac{1}{3}$, which is a contradiction. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{7}{9}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{7}{9}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in $(*)$ is no more than $\frac{7}{9}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{7}{9}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. I don't know whether the constant $\frac{7}{9}$ is optimal (for the lemma, or for the answer).

Edit: I improved the constant to $c = \frac{2}{3}$.

Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \qquad (*) $$ for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

Lemma. Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

Proof. Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$. For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

Proof of the claim. We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in $(*)$ is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

Remark. It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.