The statement to prove is the following.
Let $I$ be a non-empty subset of the real numbers which contains an interval of the form $[a,+\infty[$, for some $a\in \mathbb R$, $x_0\in \overline I$ and $f,g\colon I\to \mathbb R$ functions such that $\lim \limits_{x\to x_0}\left(g(x)\right)=+\infty$ and $\lim \limits_{x\to x_0}\left(f(x)\right)=L$, for some $L\in \mathbb R$.
In these conditions it holds true that $\lim \limits_{x\to x_0}\left((f+g)(x)=+\infty\right)$.
To prove this recall that $$\lim \limits_{x\to x_0}\left(f(x)\right)=L\iff \forall \varepsilon >0\,\exists \delta _2>0\,\forall x\in I\left(0<|x-x_0|<\delta _2\implies |f(x)-L|<\varepsilon\right)$$
and $$\lim \limits_{x\to x_0}(g(x))=+\infty\iff \forall \color{purple}M>0\,\exists \delta _1>0\,\forall x\in I(0<|x-x_0|<\delta _1\implies g(x)>\color{purple}M).$$
Remember the goal is to prove that $\lim \limits_{x\to x_0}\left((f+g)(x)=+\infty\right)$ or equivalently $$\forall \color{blue}M>0\,\exists \delta_*>0\,\forall x\in I(0<|x-x_0|<\delta _*\implies (f+g)(x)>\color{blue}M).$$
Proof: Begin by taking an arbitrary (blue) $\color{blue}M>0$.
Either $\color{blue}M\leq L-1$ holds or $\color{blue}M>L-1$ does.
$\bbox[5px,border:2px solid #000000]{\text{Case: }\color{blue}M> L-1}$
The goal is to find $\delta _*>0$ such that $\forall x\in I(0<|x-x_0|<\delta _*\implies g(x)>\color{blue}M)$.
With $\varepsilon=1$ one gets the existence of $\delta _2>0$ with the property that $\forall x\in I(0<|x-x_0|<\delta _2\implies |f(x)-L|<1)$.
With $\color{purple}M=\color{blue}M-(L-1)\color{grey}{>0}$. one gets the existence of $\delta _1>0$ such that $\forall x\in I(0<|x-x_0|<\delta _1\implies g(x)>\color{blue}M-(L-1))$.
Now define $\delta _*:=\min\left(\{\delta _1, \delta _2\}\right)$. The goal is now to prove that $\forall x\in I(0<|x-x_0|<\delta _*\implies (f+g)(x)>\color{blue}M)$.
Take $x\in I$ and assume that $0<|x-x_0|<\delta _*$.
Since $\delta _*\leq \delta _2$ one gets $|f(x)-L|<1$, i.e., $-1<f(x)-L<1$ which implies $L-1<f(x)$.
Since $\delta _*\leq \delta _1$ one gets $\color{blue}M-(L-1)<g(x)$.
Therefore $L-1+\color{blue}M-(L-1)<f(x)+g(x)$, that is, $(f+g)(x)>\color{blue}M$.
$\bbox[5px,border:2px solid #000000]{\text{Case: }\color{blue}M\leq L-1}$
Hopefully, after reading the case above, you can get an idea to solve this case.
Best Answer
Spivak's proof can be rewritten so that it is no longer a proof by contradiction.
Proof Given $\varepsilon >0$; pick $\delta >0$ such that $|f(x)-\ell|$ and $|f(x)-m|$ are both less then $\varepsilon/2$ for $0<|x-a|<\delta$; exactly as he did, take a minimum of two appropriate deltas. Now we have that $$|\ell-m|\leqslant |\ell-f(x)|+|f(x)-m|<\frac \varepsilon 2+\frac \varepsilon 2=\varepsilon$$
Since $|\ell-m|\geqslant 0$, and $|\ell-m|<\varepsilon$ is true (as we just showed) for any arbitrary $\varepsilon >0$, we see that $\ell=m$. $\blacktriangleleft$
Proof (Contrapositive) Suppose $x>0$. Then $x>\frac x 2>0$, so taking $\varepsilon =x/2$ we see that for some $\varepsilon >0$, $x\geqslant \varepsilon$. $\blacktriangleleft$
Proof By the claim, $x\leqslant 0$. Since we already have $x\geqslant 0$, $x=0$. $\blacktriangleleft$