Proving: $\partial_v f(a)$ exists if and only if so does $\partial_v f_k(a)$ $(\forall k)$ (in Normed Vector Spaces). I'm rereading some sections of Spivak's Calculus on Manifolds and attempting to generalize some results (usually to Normed Vector Spaces). I'm struggling to prove the (generelized version of) the theorem below, so I thought of asking for a hand. Throughout the post let $V$ and $W$ be NVSs, with $V'\subseteq V$ open and $f$ a function $V'\to W$.
Definition: given $a\in V'$ and a vector $v\in V$, we define (assuming the limit exists)
$$\frac{\partial f}{\partial v}(a) := \lim_{t\rightarrow 0}\frac{f(a+tv)-f(a)}{t},$$
and, in the case $v=e_i$ for a basis vector $e_i$,
$$\frac{\partial f}{\partial x_i}(a) := \frac{\partial f}{\partial e_i}(a).$$

Theorem: let $\dim W = m$ so that $f(v)=(f_1(v), \ldots ,f_m(v))$. Then $\partial_v f(a)$ exists if and only if so does $\partial_v f_k(a)$ $(\forall k)$, in which case
$$\frac{\partial f}{\partial v}(a) = \sum_{k=1}^me_k\frac{\partial f_k}{\partial v}(a).$$
Proof:
\begin{equation}
\begin{split}
\frac{\partial f}{\partial v}(a) & = \lim_{h\to 0}\frac{f(a+hv)-f(a)}{h}\\
& = \lim_{h\to 0}\left(\sum_{k=1}^me_k\frac{f_k(a+hv)-f_k(a)}{h}\right)\\
& \stackrel{*}{=} \sum_{k=1}^me_k\left(\lim_{h\to 0}\frac{f_k(a+hv)-f_k(a)}{h}\right)\\
& = \sum_{k=1}^me_k\frac{\partial f_k}{\partial v}(a).\\
\end{split}
\end{equation}

In my attempt to justify the equality $\stackrel{*}{=}$, I've come up with -what I believe is- a lemma which if true would legitimize the equality. However, now I find myself struggling to prove said lemma.
Is the lemma below true? If so, how could one prove it? If not, how else could one justify the equality $\stackrel{*}{=}$?
Lemma (?): Let $X$ be an arbitrary topological space with $p$ a limit point of $X'\subseteq X$. Let $W$ be a topological vector space with $\{e_\alpha\}_{\alpha\in A}$ as a basis of $W$, and let $f:X'\to W$ be a function. Then
$$\lim_{x\to p} f(x) = L \iff \lim_{x\to p} f_\alpha(x) = L_\alpha \ (\forall \alpha\in A)$$
where by $f_\alpha$ (or $L_\alpha$) I mean the $\alpha$ component of $f$ (or $L$).
 A: Your lemma is not true for an arbitrary topological vector space.  In an arbitrary topological vector space $W$ with basis $(e_\alpha)$, the function $p_\alpha:W\to\mathbb{R}$ taking a vector to its $\alpha$th component need not be continuous (so the forward direction of your statement will fail).  Indeed, any nonzero linear functional $W\to\mathbb{R}$ can be realized as $p_\alpha$ with respect to some basis, and a linear functional does not have to be continuous.
It is true for finite-dimensional topological vector spaces.  In this case, you can just use the fact that any finite-dimensional topological vector space is isomorphic (as a topological vector space) to $\mathbb{R}^n$ with the product topology for some $n$.  Then your statement just amounts to the fact that the change-of-basis map $T:\mathbb{R}^n\to\mathbb{R}^n$ between the standard basis and the chosen basis $(e_\alpha)$ is a homeomorphism (which is true since any linear map $\mathbb{R}^n\to\mathbb{R}^n$ is continuous, so both $T$ and $T^{-1}$ are continuous).
