Proving that a subset of $C[a,b]$ is of the first category I'm trying to prove that the set $X$ of all $f \in C[a,b]$ which are differentiable at some fixed point $c \in (a,b)$ is of the first category in $C[a,b]$ (i.e. a countable union of nowhere dense sets). 
My initial idea was to consider the sets
$$ X_n = \{ f \in C[a,b] : \left|\frac{f(x)-f(c)}{x-c}\right| \leq n, \; x\in [a,b]\}, $$
which should satisfy $X = \bigcup_{n \geq 1} X_n$, and prove that each one of them is nowhere dense. However, I am unsure how to do this or if this is even true.
Could someone help me with this?
I would prefer to just get a hint rather than a complete answer if possible.
EDIT: As pointed out in the comment section, the sets $X_n$ presented above contain functions which are not differentiable at $c$. Instead, we could consider 
$$ X_n = \{ f \in C[a,b] : \left|f'(c)\right| \leq n\}. $$
 A: For $n\in\mathbb{N}$ large enough consider
$$A_n = \left\{f \in C[a,b] :  0 < |h| < \frac1n  \implies \left|\frac{f(c+h) - f(c)}{h}\right| \le n\right\}$$
$A_n$ is closed in $C[a,b]$. Indeed, if $(f_j)_j$ is a sequence in $A_n$ such that $f_j \to f$ uniformly, then for $0 < |h| < \frac1n$ we have $\left|\frac{f_j(c+h) - f_j(c)}{h}\right| \le n$. Letting $j \to \infty$ gives $\left|\frac{f(c+h) - f(c)}{h}\right| \le n$ so $f \in A_n$.
$A_n$ is nowhere dense in $C[a,b]$. Indeed, take $\varepsilon > 0$ small enough and $f \in A_n$ be arbitrary. Define $g \in C[a,b]$ as
$$g(x) = \begin{cases} 0, &\text{ if } x \in \left[a, c-\frac{\varepsilon}{6n}\right] \\
 \frac32n\left(x + c - \frac{\varepsilon}{6n}\right), &\text{ if } x \in \left[c-\frac{\varepsilon}{6n}, c+\frac{\varepsilon}{6n} \right] \\
 \frac\varepsilon2, &\text{ if } x \in \left[c+\frac{\varepsilon}{6n}, b\right] \\
\end{cases}$$
and set $f_1 = f + g$. We have $\|f_1 - f\|_\infty = \|g\|_\infty = \frac\varepsilon2$. For $|h| <  \frac{\varepsilon}{6n}$ we have $\left|\frac{g(c+h) 
- g(c)}{h}\right| =3n$ so 
$$\left|\frac{f_1(c+h) - f_1(c)}h\right| \ge \left|\frac{g(c+h) 
- g(c)}{h}\right| - \left|\frac{f(c+h) 
- f(c)}{h}\right| \ge 3n - n = 2n > n$$
Therefore $f_1 \notin A_n$ so $B(f, \varepsilon) \not\subseteq A_n$. Since $f$ and $\varepsilon$ were arbitrary, we conclude $\operatorname{Int} A_n = \emptyset$ so $A_n$ is nowhere dense.

Let $f \in C[a,b]$ be differentiable at $c$. Then $\exists \delta > 0$ such that $0 < |h| < \delta \implies \left|\frac{f(c+h) - f(c)}h\right| \le 1 + |f'(c)|$. Take $n\in\mathbb{N}$ such that $\frac1n < \delta$ and $1 + |f'(c)| < n$. Then $f \in A_n$.
Therefore $$\{f \in C[a,b] : f \text{ is differentiable at } c\} \subseteq \bigcup_{n \in \mathbb{N}} A_n$$
is of first category in $C[a,b]$.
