If $E$ is the strict inductive limit of Frechet spaces $\lbrace E_k\rbrace$, why can't the countable basis of $E_1$ be appropriated for $E$?

Question

Not wishing to dispute that strict LF-spaces are non-metrisable, I am trying to see the flaw in the intuitive sense I have that we could appropriate, for a strict LF space, a countable basis from the smallest of the Frechet spaces in its inductive sequence. I will develop this, and would appreciate some advice on where the flaw lies.

1. Let $i, k\in\mathbb{N}$, $E$ be the strict inductive limit of the distinct Frechet spaces $\lbrace E_k\rbrace$, $\lbrace \phi_k\rbrace$ the respective natural injection isomorphisms of $E_1$ into $\lbrace E_k\rbrace$, and $\lbrace U_i\rbrace$ a basis of neighbourhoods of zero in $E_1$.
2. Since $\phi_k$ is an isomorphism, $\phi_k^{-1}$ is continuous. Hence, the preimage of any $U_i$ under $\phi_k^{-1}$ is a neighbourhood of zero in $E_k$. Therefore, $\lbrace \phi_k(U_i)\rbrace$ is a sequence of neighbourhoods of zero in $E_k$.
3. Let $V_k$ be a neighbourhood of zero in $E_k$. Since $E_1$ is a neighbourhood in $E_k$ due to the continuity of $\phi_k^{-1}$ as before, $V_k\cap E_1$ is a neighbourhood of zero in $E_k$ contained in $E_1$, and is consequently a neighbourhood in the subspace $\phi_k(E_1)$ of $E_k$ with the induced topology. Hence, $\phi_k(U_i)\subset V_k\cap E_k$ for some $i$. A fortiori, $\phi_k(U_i)\subset V_k$.
4. By 2 and 3, $\lbrace \phi_k(U_i)\rbrace$ is a countable basis of neighbourhoods of zero in $E_k$.
5. Let $V$ be a neighbourhood of zero in $E$. Then $V_k=V\cap E_k$ is a neighbourhood of zero in $E_k$ by the definition of an LF space. Hence, in terms of sets, $U_i\subset V_k\subset V$. It remains to show that $U_i$ is a neighbourhood of zero in $E$.
6. $U_i\cap E_k$ is the set identical to $\phi_k(U_i)\cap E_k$ which, since $\phi_k(U_i)\subset\phi_k(E_1)\subset E_k$, is identical to $\phi_k(U_i)$. This, in turn, is a neighbourhood of zero in $E_k$, using 2. Since the restriction of $U_i$ to $E_k$ is a neighbourhood of zero in $E_k$, $U_i$ is immediately a neighbourhood of zero in $E$.
7. By 5 and 6, $\lbrace U_i\rbrace$ is a countable basis of neighbourhoods of zero in $E$. In other words, $\lbrace U_i\rbrace$ is such a basis because each $U_i$ is a neighbourhood of zero in $E$ (point 6), any open neighbourhood $V$ in $E$ contains some $U_i$ (point 5), and $i$ runs over $\mathbb{N}$.

Since the above would make the strict LF-space $E$ metrisable, there is clearly a problem in my reasoning somewhere.