Continuity at $c=0$ is continuity from the right, so one wants to show given $\varepsilon$ there is $\delta$ for which $0 \le x < \delta$ implies $0 \le \sqrt{x} < \varepsilon$. The natural choice here is $\delta=\varepsilon^2.$
Otherwise $c>0$ and we initially restrict to $\delta<c/2$ (and throw that in at the end via the usual "min" definition.) From this assumption we have $c/2<x<3c/2$ and so
$$\sqrt{c/2}<\sqrt{x}<\sqrt{3c/2}.$$
Now we consider that
$$|\sqrt{x}-\sqrt{c}|=\frac{|x-c|}{\sqrt{x}+\sqrt{c}}. \tag{1}$$
Given our bounds for $\sqrt{x}$ the right side of $(1)$ is at most $|x-c|/(\sqrt{c/2}+\sqrt{c}),$ so that if we finally define $\delta$ as the minimum of the two numbers $c/2$ and $\varepsilon \cdot (\sqrt{c/2}+\sqrt{c})$ we'll have $|x-c|<\delta$ implies $|\sqrt{x}-\sqrt{c}|<\varepsilon.$
To show that $f(x) = 5x + 7$, no your proof will not suffice. Primarily, the step where you go from
$$\lim_{x \to a} 5x + 7 = 5a + 7.$$
Why is this true? If I make a step like this, the justification is typically because of continuity, but continuity is what you're trying to prove. You need to use the definition of continuity, which involves the metric. That is, you need to show that, for all $\varepsilon > 0$, there exists a $\delta$ such that
$$0 < D(x, a) < \delta \implies D(f(x), f(a)) < \varepsilon,$$
or, substituting in the particular metric and function,
$$0 < |x - a| < \delta \implies |5x + 7 - (5a + 7)| < \varepsilon.$$
Note the role that the metric plays in this. Changing the metric could easily make this function not continuous, and thus the limit of $5x + 7$ may not be $5a + 7$. It's your job to show that, for this particular metric, the function is continuous. That is, you have to find this $\delta$, for the given $\varepsilon$. I'll leave you to complete the proof.
For the two variable function, we do much the same thing, but we have to bear in mind that the domain is different, and hence so is the metric. I'm going to call $D_2$ the Euclidean metric on $\mathbb{R}^2$. Fix a point $(a, b) \in \mathbb{R}^2$. Then, for all $\varepsilon > 0$, we need a $\delta$ such that
$$0 < D_2((x, y), (a, b)) < \delta \implies |f(x, y) - f(a, b)| < \varepsilon,$$
or equivalently,
$$0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \implies |x + y - a - b| < \varepsilon.$$
This one isn't as straightforward as the previous one. We need a few tricks. For one, note that
$$|x - a| = \sqrt{(x - a)^2} \le \sqrt{(x - a)^2 + (y - b)^2},$$
and similarly for $|y - b|$. So, forcing the above expression to be less than $\delta$ implies both $|x - a| < \delta$ and $|y - b| < \delta$.
For two, we can use the triangle inequality (for the absolute value metric on $\mathbb{R}$) to show that $|x + y - a - b| \le |x - a| + |y - b|$. So, if we force $\delta = \frac{\varepsilon}{2}$, we should get the result we need (try filling in the blanks yourself).
As for your third question, we again use the same definition. Fix a function $h \in X$, and an $\varepsilon > 0$. We wish to find a $\delta$ such that
$$0 < D(g, h) < \delta \implies |f(g) - f(h)| < \varepsilon.$$
(Note that I substituted in the metric for $\mathbb{R}$ already.) That is,
$$0 < \sup_{x \in [0, 1]} |g(x) - h(x)| < \delta \implies |g(0) - h(0)| < \varepsilon.$$
But, clearly $|g(0) - h(0)|$ lies in the set of numbers over which we are taking the supremum, so
$$|g(0) - h(0)| \le \sup_{x \in [0, 1]} |g(x) - h(x)|.$$
If we take $\delta = \varepsilon$, then we are done!
Best Answer
The function $h$ is a product of two continuous function. What about the expoonential factor? This is a composition of two continuous function: an exponential one and the sine. The composition of continuous functions is continuous. Try to handle $g$ in tha similar manner.