Suppose that we are given jobs $$ J=(j_1,j_2,j_3,...,j_w)$$ EFT selects jobs $$ S:=(s_1,s_2,\dots <!--\; \dots \; -->s_k)$$. Recall that an optimal solution is one that maximizes the number of scheduled jobs, suppose the optimal solutions all have length $$ m$$. Before continuing we should note that $$ k<<!--\; <\; -->m$$, this is because $$ S$$ is not optimal

Let $$ \mathrm{opt}<!--\; \; -->\left(J\right)$$ be the collection of optimal solutions (note that it is finite), and for each $$ O\in <!--\; \in \; -->\mathrm{opt}<!--\; \; -->\left(J\right)$$, there is some $$ l\in <!--\; \in \; -->[1\dots <!--\; \dots \; -->k]$$ such that for every $$ j\in <!--\; \in \; -->[1\dots <!--\; \dots \; -->l],s_j=o_j$$ and $$ s_{l+1}\ne <!--\; \ne \; -->o_{l+1}$$, we will denote this value by $$ M_O$$. Note that this formalizes the idea of two tuples matching up to a certain index.

The above defines $$ f:\mathrm{opt}<!--\; \; -->\left(J\right)\to <!--\; \to \; -->{\mathbb{N}}_0$$ as $$ f\left(O\right):=M_O$$. In english language, $$ f$$ maps an optimal solution, to the index that it matches up to with respect to EFT's solution. Let $$ r:=max\left(f(\mathrm{opt}<!--\; \; -->\left(J\right))\right)$$, then let $$ O^{\star <!--\; \star \; -->}\in <!--\; \in \; -->f^{-<!--\; -\; -->1}\left(r\right)$$ be an optimal solution that matches with $$ S_{\mathrm{E}\mathrm{F}\mathrm{T}}$$ for the longest.

It's possible that $$ r=k$$, if that's true then we know that for every $$ i\in <!--\; \in \; -->[1\dots <!--\; \dots \; -->k],s_i=o_i$$. Let $$ m\ge <!--\; \ge \; -->j><!--\; >\; -->k$$, then note that $$ o_j$$ has the property that $$ S\cup <!--\; \cup \; -->\left\{o_j\right\}$$ is compatible, therefore since $$ o_j\in <!--\; \in \; -->J\setminus <!--\; \setminus \; -->S$$, then by the lemma, EFT would have extended it's solution by one on the iteration after it added $$ s_k$$ producing a solution on the next iteration of $$ S=(s_1,\dots <!--\; \dots \; -->,s_k,x)$$ where $$ x\in <!--\; \in \; -->J$$, but clearly now we have $$ S\ne <!--\; \ne \; -->S$$, which is a contradiction, thus we know $$ r\ne <!--\; \ne \; -->k$$

We are sure that $$ k<<!--\; <\; -->r$$ is not true because $$ S$$ only has $$ k$$ indices. So for sure $$ r<<!--\; <\; -->k$$, by the way $$ r$$ was defined this means that for every $$ j\in <!--\; \in \; -->[1\dots <!--\; \dots \; -->r],s_j=o_j$$, but that $$ s_{r+1}\ne <!--\; \ne \; -->o_{r+1}$$. Note that since $$ s_{r+1}\ne <!--\; \ne \; -->o_{r+1}$$ and ETF adds by earliest finishing time, we must have that $$ f_{s_{r+1}}\le <!--\; \le \; -->f_{o_{r+1}}$$, since that's true we know that $$ \{s_{r+1},o_{r+2},\dots <!--\; \dots \; -->o_m\}$$ is compatible, therefore $$ (o_1,o_2,\dots <!--\; \dots \; -->,o_r,s_{r+1},\dots <!--\; \dots \; -->,o_m)$$ is an optimal solution that matches with $$ S$$ for $$ r+1$$ indices. This is impossible, because the maximum number of indices an optimal solution could match was $$ r$$.

Our assumption was that ETF doesn't produce an optimal solution, and we obtained a contradiction.