$$\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)$$
Since $f$ and $g$ are continuous, $|f(x)-f(a)|< ε_1$ when $0<|x-a|< \delta_f $ and $|g(x)-g(a)|< ε_2$ $\ $ when $0<|x-a|< _g$
Let $$ be defined as $\min(_g ,_f)$ and $h(x)$ be defined as $f(x)+g(x)$
$\ |f(x)-f(a)|+|g(x)-g(a)|< ε_1+ε_2$ (1)
$\ |f(x)-f(a) + g(x)-g(a)|< ε_1+ε_2$
$\ |h(x)-h(a)|< ε_1+ε_2$
We can replace $_f$ and $_g$ by $$ to get
$|f(x)-f(a)| < ε_1$ when $0<|x-a|<$ and $\ |g(x)-g(a)|< ε_2$ when $0<|x-a|<$
Since (1) is true when $0<|x-a|<$, we can find a $$ for every value of $ε_1+ε_2$
$\ |h(x)-h(a)|< ε_1+ε_2$ can be rewritten as $\ |h(x)-h(a)|< r_1+r_2$
Any real number can be expressed as the sum of two real numbers so r=$\ r_1+r_2$
Which gives us $\ |h(x)-h(a)|< r$ when $0<|x-a|< $
substituting r=ε we get $\ |h(x)-h(a)|< ε$ when $0<|x-a|< $ for all positive real values of ε.
Is this an acceptable proof?