The simplest explanation is that it is a mistake. One can however complete the proof as follows:

If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of distinct primes in $L$ dividing $m$ is at least $u(m)$, unless $u(m) = 2$ and $m$ has only one prime factor in $L$, in which case $m = p^2 q$.

Let us count pairs of integers $(p,q)$ where $p > R := c_{8} X^{\frac{1}{e}}$ and $m < p^2 q \leq m+t$. This it at most $$ \sum_{q \leq X R^{-2}} \mathrm{number \ of \ squares \ in \ } ]mq^{-1},(m+t)q^{-1} ] \\ \leq \sum_{q \leq X R^{-2}} \left( \frac{t}{\sqrt{qm}} + O(1) \right) \\ \ll \frac{t}{\sqrt{R}} \sqrt{X R^{-2}} + X R^{-2} \\ \ll t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha} $$ where $\alpha = \frac{3-e}{2e} >0$.

Thus the number of distinct prime factors of $\prod_{j=1}^{t} (m+j)$ in $L$ is at least $t - O(t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha})$. By choosing $t$ of size $X^{\frac{1}{e}}$, one thus gets $\geq t - O(t X^{-\alpha})$ distinct prime factors $>t$. By adding the $\pi(t)$ primes $\leq t$, we get $\geq t$ distinct prime factors (for $X$ large enough).

EDIT: As pointed out Gerhard Paseman below, I answered a different question ... The original question can be answered as follows :

Let $\omega_{>k}(n)$ be the number of distinct prime factors $>R$ of $n$. We first note that $$ \nu(n,k) = \pi(k) + \sum_{i=1}^k \omega_{>k}(n+i). $$ Let $c >0$ be large enough so that $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \leq \left( 1 - \frac{3e}{\log X} \right) X $$ holds for large $X$ with $R = c X^{\frac{1}{e}}$ (indeed the inequality holds with $3e$ replaced by $e \log c + O(1)$). We first show that for any $X$ large enough the following holds:

$(*)$ there exists a $n \in [X,2X]$ such that for each $k \in [R, X^{\frac{4}{5}}]$, one has $\nu(n,k) < k$.

Indeed, assume the contrary for some $X$. Then starting with $n_0 =X$, we get a $k_0 \in [R, X^{\frac{4}{5}}]$ such that $\nu(n_0,k_0) \geq k_0$, and then with $n_1 = n_0 + k_0$ some $k_1$ such that $\nu(n_1,k_1) \geq k_1$, .... and so on until $n_{J+1} = n_J + k_J > 2X$ for some $J$. We then have $$ \sum_{n_j < n \leq n_{j+1}} \omega_{>R}(n) \geq \sum_{n_j < n \leq n_{j+1}} \omega_{>k_j}(n) \geq k_j - \pi(k_j) \geq \left( 1 - \frac{2}{\log R} \right) k_j. $$ Summing over $j$, this yields $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \geq \left( 1 - \frac{2}{\log R} \right) X,$$ which contradicts our choice of $c$ for $X$ large enough.

Thus for any $X$ large enough one can take $n \in [X,2X]$ as in $(*)$. One has $\nu(n,k) < k$ for $k \in [R, X^{\frac{4}{5}}]$. But for $k > X^{\frac{4}{5}}$, a direct count using Brun-Titschmarsh inequality yields $\nu(n,k) \leq c k + o(k)$ with $c = 2 \log \frac{4}{3} < 1$, hence $\nu(n,k) < k$ when $X$ is large enough. Thus $f_0(n) < R \leq c n^{\frac{1}{e}}$.

The simplest explanation is that it is a mistake. One can however complete the proof as follows:

If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of distinct primes in $L$ dividing $m$ is at least $u(m)$, unless $u(m) = 2$ and $m$ has only one prime factor in $L$, in which case $m = p^2 q$.

Let us count pairs of integers $(p,q)$ where $p > R := c_{8} X^{\frac{1}{e}}$ and $m < p^2 q \leq m+t$. This it at most $$ \sum_{q \leq X R^{-2}} \mathrm{number \ of \ squares \ in \ } ]mq^{-1},(m+t)q^{-1} ] \\ \leq \sum_{q \leq X R^{-2}} \left( \frac{t}{\sqrt{qm}} + O(1) \right) \\ \ll \frac{t}{\sqrt{R}} \sqrt{X R^{-2}} + X R^{-2} \\ \ll t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha} $$ where $\alpha = \frac{3-e}{2e} >0$.

Thus the number of distinct prime factors of $\prod_{j=1}^{t} (m+j)$ in $L$ is at least $t - O(t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha})$. By choosing $t$ of size $X^{\frac{1}{e}}$, one thus gets $\geq t - O(t X^{-\alpha})$ distinct prime factors $>t$. By adding the $\pi(t)$ primes $\leq t$, we get $\geq t$ distinct prime factors (for $X$ large enough).

EDIT: As pointed out Gerhard Paseman below, I answered a different question ... The original question can be answered as follows :

Let $\omega_{>k}(n)$ be the number of distinct prime factors $>k$ of $n$. We first note that $$ \nu(n,k) = \pi(k) + \sum_{i=1}^k \omega_{>k}(n+i). $$ Let $c >0$ be large enough so that $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \leq \left( 1 - \frac{3e}{\log X} \right) X $$ holds for large $X$ with $R = c X^{\frac{1}{e}}$ (indeed the inequality holds with $3e$ replaced by $e \log c + O(1)$). We first show that for any $X$ large enough the following holds:

$(*)$ there exists a $n \in [X,2X]$ such that for each $k \in [R, X^{\frac{4}{5}}]$, one has $\nu(n,k) < k$.

Indeed, assume the contrary for some $X$. Then starting with $n_0 =X$, we get a $k_0 \in [R, X^{\frac{4}{5}}]$ such that $\nu(n_0,k_0) \geq k_0$, and then with $n_1 = n_0 + k_0$ some $k_1$ such that $\nu(n_1,k_1) \geq k_1$, .... and so on until $n_{J+1} = n_J + k_J > 2X$ for some $J$. We then have $$ \sum_{n_j < n \leq n_{j+1}} \omega_{>R}(n) \geq \sum_{n_j < n \leq n_{j+1}} \omega_{>k_j}(n) \geq k_j - \pi(k_j) \geq \left( 1 - \frac{2}{\log R} \right) k_j. $$ Summing over $j$, this yields $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \geq \left( 1 - \frac{2}{\log R} \right) X,$$ which contradicts our choice of $c$ for $X$ large enough.

Thus for any $X$ large enough one can take $n \in [X,2X]$ as in $(*)$. One has $\nu(n,k) < k$ for $k \in [R, X^{\frac{4}{5}}]$. But for $k > X^{\frac{4}{5}}$, a direct count using Brun-Titschmarsh inequality yields $\nu(n,k) \leq c k + o(k)$ with $c = 2 \log \frac{4}{3} < 1$, hence $\nu(n,k) < k$ when $X$ is large enough. Thus $f_0(n) < R \leq c n^{\frac{1}{e}}$.

The simplest explanation is that it is a mistake. One can however complete the proof as follows:

If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of distinct primes in $L$ dividing $m$ is at least $u(m)$, unless $u(m) = 2$ and $m$ has only one prime factor in $L$, in which case $m = p^2 q$.

Let us count pairs of integers $(p,q)$ where $p > R := c_{8} X^{\frac{1}{e}}$ and $m < p^2 q \leq m+t$. This it at most $$ \sum_{q \leq X R^{-2}} \mathrm{number \ of \ squares \ in \ } ]mq^{-1},(m+t)q^{-1} ] \\ \leq \sum_{q \leq X R^{-2}} \left( \frac{t}{\sqrt{qm}} + O(1) \right) \\ \ll \frac{t}{\sqrt{R}} \sqrt{X R^{-2}} + X R^{-2} \\ \ll t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha} $$ where $\alpha = \frac{3-e}{2e} >0$.

Thus the number of distinct prime factors of $\prod_{j=1}^{t} (m+j)$ in $L$ is at least $t - O(t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha})$. By choosing $t$ of size $X^{\frac{1}{e}}$, one thus gets $\geq t - O(t X^{-\alpha})$ distinct prime factors $>t$. By adding the $\pi(t)$ primes $\leq t$, we get $\geq t$ distinct prime factors (for $X$ large enough).

The simplest explanation is that it is a mistake. One can however complete the proof as follows:

If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of distinct primes in $L$ dividing $m$ is at least $u(m)$, unless $u(m) = 2$ and $m$ has only one prime factor in $L$, in which case $m = p^2 q$.

Let us count pairs of integers $(p,q)$ where $p > R := c_{8} X^{\frac{1}{e}}$ and $m < p^2 q \leq m+t$. This it at most $$ \sum_{q \leq X R^{-2}} \mathrm{number \ of \ squares \ in \ } ]mq^{-1},(m+t)q^{-1} ] \\ \leq \sum_{q \leq X R^{-2}} \left( \frac{t}{\sqrt{qm}} + O(1) \right) \\ \ll \frac{t}{\sqrt{R}} \sqrt{X R^{-2}} + X R^{-2} \\ \ll t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha} $$ where $\alpha = \frac{3-e}{2e} >0$.

Thus the number of distinct prime factors of $\prod_{j=1}^{t} (m+j)$ in $L$ is at least $t - O(t X^{-\alpha} +X^{\frac{1}{e} - 2 \alpha})$. By choosing $t$ of size $X^{\frac{1}{e}}$, one thus gets $\geq t - O(t X^{-\alpha})$ distinct prime factors $>t$. By adding the $\pi(t)$ primes $\leq t$, we get $\geq t$ distinct prime factors (for $X$ large enough).

EDIT: As pointed out Gerhard Paseman below, I answered a different question ... The original question can be answered as follows :

Let $\omega_{>k}(n)$ be the number of distinct prime factors $>R$ of $n$. We first note that $$ \nu(n,k) = \pi(k) + \sum_{i=1}^k \omega_{>k}(n+i). $$ Let $c >0$ be large enough so that $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \leq \left( 1 - \frac{3e}{\log X} \right) X $$ holds for large $X$ with $R = c X^{\frac{1}{e}}$ (indeed the inequality holds with $3e$ replaced by $e \log c + O(1)$). We first show that for any $X$ large enough the following holds:

$(*)$ there exists a $n \in [X,2X]$ such that for each $k \in [R, X^{\frac{4}{5}}]$, one has $\nu(n,k) < k$.

Indeed, assume the contrary for some $X$. Then starting with $n_0 =X$, we get a $k_0 \in [R, X^{\frac{4}{5}}]$ such that $\nu(n_0,k_0) \geq k_0$, and then with $n_1 = n_0 + k_0$ some $k_1$ such that $\nu(n_1,k_1) \geq k_1$, .... and so on until $n_{J+1} = n_J + k_J > 2X$ for some $J$. We then have $$ \sum_{n_j < n \leq n_{j+1}} \omega_{>R}(n) \geq \sum_{n_j < n \leq n_{j+1}} \omega_{>k_j}(n) \geq k_j - \pi(k_j) \geq \left( 1 - \frac{2}{\log R} \right) k_j. $$ Summing over $j$, this yields $$ \sum_{X<n \leq 2X + X^{\frac{4}{5}}} \omega_{>R}(n) \geq \left( 1 - \frac{2}{\log R} \right) X,$$ which contradicts our choice of $c$ for $X$ large enough.

Thus for any $X$ large enough one can take $n \in [X,2X]$ as in $(*)$. One has $\nu(n,k) < k$ for $k \in [R, X^{\frac{4}{5}}]$. But for $k > X^{\frac{4}{5}}$, a direct count using Brun-Titschmarsh inequality yields $\nu(n,k) \leq c k + o(k)$ with $c = 2 \log \frac{4}{3} < 1$, hence $\nu(n,k) < k$ when $X$ is large enough. Thus $f_0(n) < R \leq c n^{\frac{1}{e}}$.