One little fact that will help resolve this is that $\dim(H)=rank(\Phi)$, the rank of the root system.
Now the rest is done by a look at the classification of root systems. If we don't have that ready, we can even get by with just the low-dimension / low-rank cases by hand:
A root system of rank $\ge 3$ must contain at least six roots (a basis and their negatives), so in this case we would already have $rank(\Phi)+\lvert \Phi\rvert \ge 9$ (usually much larger in fact, but dimension $9$ indeed occurs for $\mathfrak{sl}_2\oplus \mathfrak{sl}_2 \oplus \mathfrak{sl}_2$).
Root systems of rank $2$ are implicitly classified at the beginning of every lecture on root system when one discusses the relations which two roots can have to each other. Turns out the possibilities are $A_1 \times A_1, A_2, B_2=C_2$, and $G_2$; while the first indeed contains four roots and describes the semisimple, six-dimensional $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$, all the others contain $\ge 6$ roots and thus correspond to Lie algebras of dimension $\ge 2+6 =8$ (actually, $A_2$ corresponds to the $8$-dimensional $\mathfrak{sl}_3$, and $B_2=C_2$ to the $10$-dimensional $\mathfrak{so}_5 \simeq \mathfrak{sp}_4$; the dimension of the exceptional Lie algebra of type $G_2$ is $14=2+12$).
There's only one root system of rank $1$: $A_1$, which corresponds to the $3$-dimensional $\mathfrak{sl}_2$. So there's nothing that could make up a semisimple Lie algebra of dimension $4,5,$ or $7$ (or dimension $1$ or $2$, for that matter; I'd think the next non-occurring dimension is $11$, edit: As Jason DeVito and Dietrich Burde point out in comments (thanks!), actually all higher dimensions do occur.)
The above implicitly assumed that we work over an algebraically closed field of characteristic $0$. As YCor points out in a comment, this is enough to conclude for any base field $k$ of characteristic $0$. Namely, if $L$ is a semisimple Lie algebra over $k$ of $k$-dimension $n$, and $K\vert k$ is any field extension, then the scalar extension $L\otimes_k K$ is a semisimple Lie algebra of $K$-dimension $n$. Apply to an algebraic closure of $k$.
Best Answer
Let $V$ be a finite-dimensional vector space with dual $V^*$. To make notation consistent (and remove the superficial differences between $V$ and its dual) it is helpful to write down a pairing $\langle -, - \rangle \colon V \times V^* \to \mathbb{C}$ defined by $\langle v, f \rangle = f(v)$.
A nonzero dual vector $f \in V^*$ defines a hyperplane $\langle -, f \rangle = 0$, or equivalently $\ker f$. Similarly, a nonzero vector $v \in V$ defines a hyperplane $\langle v, - \rangle = 0$ (sometimes called the annihilator of $v$). This generalises: for any set $S \subseteq V$ of vectors, we can define $S^\perp \subseteq V^*$ by $$ S^\perp = \{f \in V^* \mid \langle s, f \rangle = 0 \text{ for all } s \in S\} \subseteq V^*.$$ Similarly, we could take a set $T \subseteq V^*$ of dual vectors and define $$ T^\perp = \{v \in V \mid \langle v, t \rangle = 0 \text{ for all } t \in T\} \subseteq V.$$ The key properties of this complement operation are (for our purposes):
These properties can be proven by looking at a basis of $V$ together with its corresponding dual basis in $V^*$.
Now to answer your question: if $\Phi \subseteq V^*$ does not span, then $\operatorname{codim}(\operatorname{span} \Phi) = \dim \Phi^\perp \geq 1$, and so there exists a nonzero vector $v \in \Phi^\perp$ such that $\langle v, \alpha \rangle = 0$ for all $\alpha \in \Phi$.