Continuous decomposition into sum of vectors along sum of linear subspaces I'm trying to prove that Minkowsky sum of relative interiors of two convex sets is contained in the relative interior of the sum. I need the following lemma to complete the proof:
assume that $U$ and $W$ are finite dimensional subspaces of Banach space $V$.

Is it true that for all positive real $\varepsilon$ it is possible to find some positive real number $\delta$ such that for every vector $v \in U + W$ with $\|v\| \le  \delta$ it is possible to find vectors $u \in U$ and $w \in W$ such that $v = w + u $ and $\| w\|< \varepsilon$ and $\| v \| < \varepsilon$? Or in language of quantors: $$ \forall \varepsilon \in \mathbb{R}_{++} \; . \;\exists \delta \in \mathbb{R}_{++} \; : \; \forall v \in U + W : \|v\| \le \delta \; . \; \exists w \in W \; : \; \exists u \in U \; : \quad  :v = u + w \; \And \; \|w\| < \varepsilon \;  \And \; \|u\| < \varepsilon $$ 

Looking at the quantified statement it seems that the problem can be restated as the problem of existence of  function $f : W + U \to W \times U $  continuous at $0$ with $f(0) = (0,0)$, and at least locally at $0$ providing the sum property $v = w + u$ with $f(v) = (w,u)$.
I tried going with contrapositive:
Assume that there exists sequence  of vectors $v_n \in W + U$  with $\lim_{n \to \infty} v_n = 0$, such that every decomposition $v_n =w_n + u_n$ (that necessarily exists) has property that $\|w_n\| \ge \varepsilon$ and $\|u_n\| \ge \varepsilon$. Assume, first, that the sum is direct i. e. $v_n \in U \oplus W$ so we have no choice while selecting $u_n$ and $w_n$. Then
$$\lim_{n \to \infty} u_n + w_n  =0  $$
implying 
$$
 \lim_{n \to \infty} \Big\|u_n - (-w_n)\Big\| =0. 
$$
If both sequences $u_n$ and $w_n$ have same converging subsequence then their partial limits $u\in U$ and $w \in W$ (as both subspaces are closed) are and non zero and have $u = -w$, hence, $U \cap W \neq \{0\}$, a contradiction.  But if one sequence is converging for a subsequence of indices, then the other must also stabilize. Indeed, $u_n + w_n$ is Cauchy:
$$ \| v_{n_k} - v_{n_l} \|=\|u_{n_k} + w_{n_k} - u_{n_l}  - w_{n_l}\| \ge \Big| \| u_{n_k} - u_{n_l} \| - \| w_{n_k} - w_{n_l}\| \Big|.  
 $$
So if $u$ is Cauchy for some subsequence, $w$ must be also Cauchy and hence converging. This provides us with the property
$$\lim_{n \to \infty} \|u_n\| = \infty$$ and 
$$ \lim_{n \to \infty} \| - w_n\| =\lim_{n \to \infty} \| w_n\| = \infty $$
as we may assume our space to be locally compact. This must imply that $ \lim_{n \to \infty} \|u_n -(-w_n)\| \neq 0 $ as vectors of high norm in trivially intersecting vector spaces must be detached and $W \cap U = \{0\}$
. This provides a contradiction.
In case $U + W \neq U \oplus W$ we can select subspace $W'$ of $W$ such that $ W = U \cap W \oplus W'$, then $U + W = U \oplus W'$ and we achieve similar result. 
Is my reasoning correct?
Is there a simpler proof? 
 A: If I'm not mistaken, you haven't used the finite-dimensionality of $U$ and $W$.
Assume that the sum is direct, i.e. $U \cap W = \{0\}$.
Let $B = \{u_1, \ldots, u_m, w_1, \ldots, w_n\}$ be a basis for $U + W$, where $u_1, \ldots, u_m \in U$ and $w_1, \ldots, w_n \in W$.
Define a norm $\|\cdot\|_{B, \infty}$ on $U + W$ like this:
$$\left\|\sum_{i=1}^m{\alpha_i}u_i + \sum_{j=1}^n \beta_jw_j\right\|_{B, \infty} = \max\{|\alpha_1|, \ldots, |\alpha_m|, |\beta_1|, \ldots, |\beta_n|\}$$
Since $U + W$ is finite-dimensional, all norms are equivalent so there exist constants $m, M > 0$ such that:
$$m\|x\|_{B, \infty} \le \|x\| \le M\|x\|_{B, \infty}, \text{ for all }x \in U + W$$
Now let $\varepsilon > 0$ and pick $\delta = \frac{m}{M}\cdot\frac{\varepsilon}2$.
For any $v \in U + W$ such that $\|v\| \le \delta$ take $u \in U$ and $v \in W$ such that $v = u + w$. Notice that $\|u\|_{B, \infty}, \|w\|_{B, \infty} \le \|v\|_{B, \infty}$.
We have:
$$\frac{m}{M}\max\{\|u\|, \|w\|\} \le m\max\{\|u\|_{B, \infty}, \|w\|_{B, \infty}\} \le m\|v\|_{B, \infty} \le \|v\| \le \delta = \frac{m}{M}\cdot\frac{\varepsilon}2$$
Hence, $$\max\{\|u\|, \|w\|\}  \le \frac{\varepsilon}2 < \varepsilon$$
As you suggest, if $U \cap W \ne \{0\}$, then $U + W = U \oplus W'$ so you could run this argument again.
