= Solution
Let $M=\|\,|e_0|+|e_1|\,\|_\infty$. Since both <linear functionals> have <operator norm> one, for every $f\in X$ the <triangle inequality> gives
$$
|(Tf)(x)|\leq|\ell_0(f)||e_0(x)|+|\ell_1(f)||e_1(x)|
\leq\|f\|_\infty\bigl(|e_0(x)|+|e_1(x)|\bigr).
$$
Taking the <essential supremum> proves $\|T\|\leq M$.
For the reverse bound, <continuity> and the <extreme value theorem> give $x_*\in[0,1]$ with $|e_0(x_*)|+|e_1(x_*)|=M$. Choose complex $a_0,a_1$ of modulus one so that $a_je_j(x_*)=|e_j(x_*)|$, taking $a_j=1$ if $e_j(x_*)=0$. The <continuous function>
$$
f_*(x)=(1-x)a_0+xa_1
$$
has <supremum norm> one by the <triangle inequality> and its endpoint values. Since $\ell_j(f_*)=a_j$,
$$
(Tf_*)(x_*)=a_0e_0(x_*)+a_1e_1(x_*)=M.
$$
The output is continuous, so its <L-infinity norm> equals its <supremum norm>, including when $x_*$ is an endpoint. Therefore $\|T\|\geq M$, and the exact <operator norm> is
$$
\boxed{\|T\|=\|\,|e_0|+|e_1|\,\|_\infty.}
$$
This is the <norm of a two-endpoint interpolation projection>.
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