Your confusion stems from a misinterpretation of a formal statement about the limitations of perfect secrecy. Let us first recall the definition
Perfect Secrecy.
An encryption scheme $(\mathsf{Gen}, \mathsf{Enc}, \mathsf{Dec})$ with message space $\mathcal{M}$ is perfectly secret if for every probability distribution over $\mathcal{M}$, every message $m \in \mathcal{M}$, and every ciphertext $c \in \mathcal{C}$ for which $\Pr[C = c] > 0$:
$$Pr[M = m | C = c] = Pr[M = m].$$
Note that there is no mention of bits or bitstrings here and "length of the message" is not really well defined for a general message space.
The statement that is often informally paraphrased as
The key must be at least as long as the message.
therefore doesn't really make any sense here. This is where you seem to be confused. The formal statement that can be proven about perfectly secret encryption schemes is the following:
Limitation of Perfect Secrecy.
If $(\mathsf{Gen}, \mathsf{Enc}, \mathsf{Dec})$ is a perfectly secret encryption scheme with message space $\mathcal{M}$ and key space $\mathcal{K}$, then $|\mathcal{K}| \ge |\mathcal{M}|$.
Note that this is only a statement about the size of the key space! It makes no mention of any length.
Now, the most well known perfectly secret encryption scheme is the one-time pad, which is most commonly defined over bitstrings¹ with a message space $\mathcal{M}=\{0,1\}^n$ for some message length $n$ as well as a keyspace $\mathcal{M}=\{0,1\}^\ell$ for some key length $\ell$. In this specific case case $|\mathcal{K}| \ge |\mathcal{M}|$ indeed implies that $\ell \ge n$.
In your specific case however you used a different message space, in particular you defined $\mathcal{M}$ to be the set of "messages in the English language" which is something we would need to define more formally. We can define this for example as "any concatenation of words in the English language with total length at most $\ell$ characters".
In this case, clearly $|\mathcal{M}| < 2^{8\ell}$ and therefore obviously you do not need $|\mathcal{K}|\geq 2^{8\ell}$ but the definition never claimed otherwise. What you are defining here is simply not the one-time-pad, but a different encryption scheme with a different message space. And the same condition on the key space still holds. It is just that this does not necessarily correspond to an immediate lower bound on the length of the keys when encoded as bitstrings.
¹Technically the one-time pad can be defined over strings over any finite alphabet $\Sigma$ as long as we can define an order over $\Sigma$.
standard OTP key of random 0-255 value bytes
which is bluntly wrong. An OTP key has to be completely random within the "alphabet" you're using. Random bytes may be handy in the world of computers as it exactly fits the hardware expectations… but OTP can also be applied to other symbol ranges with different bit sizes. One could even apply OTP on a 0-1 symbol range (think bit-wise instead of byte-wise) as long as the series of 0s and 1s is absolutely random. Take a simple Y/N msg, and your entropy theories can be voided with a simple coin flip. $\endgroup$